What is the highest number of digits so that this number of digits in a specific power of 2 are exactly 10%?

Here's a very rough heuristic argument (this is too long for a comment).

Let's assume that the digits of a power of $2$ are roughly "random" -- that is, if the number has $\ell$ digits, each of the $\ell$ digits is chosen uniformly at random from $\{0,1,\dots,9\}$. This is clearly not exactly correct, but it's a reasonable guess for at least the middle digits, which contribute most of the "arbitrariness" to the problem.

Let $b=10$, and pick some subset $S\subset\{0,\dots,b-1\}$ of digits. We'll calculate the probability that, for a number with $bk$ digits chosen randomly, for each digit $d\in S$, the number of occurrences of $d$ is exactly $k$ (i.e. $d$ occurs at exactly $1/b$ of the digits). Each of the $b^{bk}$ $bk$-digit numbers (including with leading $0$s) are equally likely to be chosen, so we need to count the number of such numbers with exactly $k$ occurrences of each $d\in S$ and divide that by $b^{bk}$.

Say $|S|=c$, and let $S=\{d_1,\dots,d_c\}$. There are $\binom{bk}k$ ways to choose the $k$ locations of $d_1$, then $\binom{bk-k}k$ ways to choose the locations of $d_2$, et cetera. So, the number of total choices for the digits in $S$ is $$\binom{bk}k\binom{bk-k}k\cdots \binom{bk-(c-1)k}k=\frac{(bk)!}{k!^c(bk-ck)!}.$$ There are $(b-c)^{(b-c)k}$ ways to choose the remaining digits, since they must all be in $\{0,1,\dots,b-1\}\setminus S$, with no other restriction. So, the desired probability is $$p_{b,c}(k):=\frac{\frac{(bk)!}{k!^c((b-c)k)!}(b-c)^{(b-c)k}}{b^{bk}}.$$

We'll now investigate, given fixed $b$ and $c$, how this quantity grows with $k$. For this, we'll use Stirling's approximation for factorials. This approximation only works with large inputs, so we'll deal with the case of $c=b$ (where we want all of the digits to occur exactly $k$ times) separately.

For $c<b$, Stirling's approximation gives the rather ugly computation \begin{align*} p_{b,c}(k) &\sim \frac{(b-c)^{(b-c)k}}{b^{bk}}\frac{\sqrt{2\pi bk}\left(\frac{bk}e\right)^{bk}}{\left(\sqrt{2\pi k}\left(\frac ke\right)^k\right)^c\sqrt{2\pi(b-c)k}\left(\frac{(b-c)k}{e}\right)^{(b-c)k}}\\ &=\sqrt{\frac{b}{(b-c)(2\pi k)^c}}, \end{align*} where $\sim$ means that the ratio between the two quantities tends to $1$ as $k\to\infty$. For $c=b$, we have $p_{b,b-1}(k)=p_{b,b}(k)$ (since if $b-1$ of the digits occur exactly $k$ times, the last must as well), and so $$p_{b,b}(k)\sim \sqrt{\frac{b}{(2\pi k)^{b-1}}}.$$

This means that, for fixed $b$ and $c$, and for large $k$, the probability that $c$ of the $b$ digits occur exactly $k$ times in a perfectly random length-$bk$ number is about some constant times $k^{-c/2}$, or $k^{-(b-1)/2}$ if $c=b$.

Now we'll apply this heuristic to the particular problem. The number of powers of $2$ of a given length in base $10$ is either $3$ or $4$, and on average it's about $r:=\log_2(10)$. Fix $S\subset\{0,1,\dots,9\}$ of size $c$. For large $N$, the expected number of powers of $2$ $2^n$ with $n>N$ with $10k$ digits for some $k$, for which every $d\in S$ occurs in $2^n$ exactly $k$ times, is very roughly $$r\sum_{k\geq \log_{10}(2^N)/10}\frac{\sqrt{\frac{10}{10-c}}}{(2\pi k)^{c/2}}=\frac{r\sqrt{\frac{10}{10-c}}}{(2\pi)^{c/2}}\sum_{k\geq rN/10}k^{-c/2}.$$

For $c>2$, this sum is finite, and should tend to $0$ quickly as $N$ grows. In particular, for $N=10^6$ (assuming it has been checked that there are no solutions $2^n$ with $n\leq 10^6$), and $c=9$ (acting as proxy for $c=10$), the sum is about $3.6\times 10^{-23}$. So, it's exceedingly unlikely, assuming our heuristic is close enough to the truth, that there are any powers of $2$ with all of the digits perfectly distributed.

For $c=2$, the sum diverges, as the harmonic series does. So, we expect infinitely many powers of $2$ to be exactly $10k$ digits long for some $k$, and to have some two digits which occur exactly $k$ times.

Some numerics to support the argument that these heuristics are reasonable:

I checked powers of $2$ up through $2^{10^5}$. Of these, exactly $10^4$ have a length which is exactly a multiple of $10$. Two of these numbers satisfied the condition for $5$ digits appearing exactly $10\%$ of the time: these are $2^{31}$ and $2^{166}$, of lengths $10$ and $50$, respectively. Two more, $2^{30}$ and $2^{398}$, satisfied the condition for $4$ digits.

Only $21$ numbers satisfied the condition for $3$ digits. These are fairly common for small $k$, and then tend to thin out, with the largest being $2^{32685}$. According to the presented heuristic, the probability that there's another such power out there is about $0.31$ (taking the value for $c=3$ and multiplying by $\binom{10}3$ to account for the number of different choices of $S$). (Edit: $2^{112380}$ works as well!)

$114$ numbers satisfied the condition for $2$ digits. These begin fairly common, but remain somewhat common throughout: the five largest such numbers I found were $$2^{90553}, 2^{92646}, 2^{93311}, 2^{96101}, 2^{98360}.$$ The fact that these approach the upper bound $10^5$ on the computation, with no signs of stopping, supports the heuristic's judgment that there should be infinitely many solutions with $c=2$.