What is the highest power of 3 that divides a string of 3^2013 digit 3s?

The reason for being incorrect is already in comments, if you use the right expression $\dfrac{10^{3^{2013}}-1}3$ for your number you'll find it is divisible by $9$. In fact it is divisible by a higher power of $3$.

Let the number made of a string of $3^k$ threes be $S_k$. Then we can show the stronger claim that $S_k \equiv 3^{k+1} \mod {3^{k+2}}$, which is enough to conclude that the power you seek is $3^{2014}$.

The base case for an induction argument is easy with $k=0$ or $k=1$. For the induction step, note $S_{k+1} = S_k \cdot \left(10^{3^{k+1}}+10^{3^k}+1 \right)$

By the induction hypothesis, we have $S_k = a\cdot 3^{k+2}+3^{k+1}$, for some $a \in \mathbb N$. Further, obviously $10^{3^{k+1}}+10^{3^k}+1 \equiv 3 \pmod 9 $, hence $10^{3^{k+1}}+10^{3^k}+1 = 9b+3$ for some $b\in \mathbb N$.

Therefore we have the induction step $$S_{k+1} = S_k \cdot \left(10^{3^{k+1}}+10^{3^k}+1 \right) = \left(a\cdot 3^{k+2}+3^{k+1}\right)\cdot(9b+3) \equiv 3^{k+2} \pmod {3^{k+3}}$$