Finding the 2,147,483,647th prime number

I'm not quite sure why you need to know what the $2147483647$-th prime number is, but I'm going to guess it's so you know what data type to store the elements of the array in.

In that case, you only need an approximate value. The Prime Number Theorem tells us that

$$\pi(n)\approx\frac{n}{\log(n)}$$

where $\pi(n)$ is the number of prime numbers less than $n$. So we are looking for $n$ with $\pi(n)\ge2147483647$ (then there are at least $2147483647$ primes less than $n$, so in particular the first $2147483647$ primes are all smaller than $n$, and can be stored in variables of size $n$).

Let's just ignore the $\approx$ sign and treat it as an equality. This means we might accidentally get a value that's too small, but let's not worry about that.

So we want

$$\pi(n)=\frac{n}{\log(n)}=2147483647$$

Wolfram|Alpha gives an answer of $5.303\times 10^{10}$ for $n$.

Now $\log_2(5.03\times 10^{10})\approx 35.5$, so you'd need 36 bits to store that number.

$\log_{10}(5.03\times10^{10})\approx 10.7$, so you'd need $11$ digits to represent the $2147483647$-th prime number in decimal.

Just so we have an answer: As Artes has shown, the exact prime is $50685770143$. This has $11$ decimal digits. In binary, it is $101111001101000110110111100110011111_2$, which has $36$ bits.