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.
Finding palindromic primes is not very hard, but rather proving them and labeling them as Prime instead of PRP (Probable Prime) is the hard job. You can find the list of world's largest palindromic primes here
And the reason why proving most of them is time-consuming lies in odd digits in them. That is if you could express your number as $k*2^n +- 1$ then proving is much easier using several algorithms out there, but since palindromic primes usually have more odd digits than $0$ ones, then their finding procedure becomes too long.
As an example, the best palindromic number is of the following form (the smallest one is $101$):
$100000000000...00000000000000001$
It can be easily expressed as $5^n*2^n + 1$ and therefore easy to do primility test. Occurance of a non-zero digit at any position, cause the power of $2$ in the formula to be decreased.
So the best place for such a digits is near the center of number.
Again as an example where power of $2$ is near $n/2$
$100000000000...007700...00000000000000001$
Updated
And for generating the number you can build a simple formula like the following in Mathematica:
Palindromic[n_, digit_, position_] :=
(10^n - 1)/9 + (digit - 1) (10^position + 10^(n - position - 1))
In[1]:= Palindromic[20, 7, 9]
Out[1]= 11111111177111111111
Best Answer
In case the exact number helps, Mathematica can compute
PrimePi[2^16] - PrimePi[2^15 - 1]to be $3030$. Choosing one-hundred odd integers uniformly at random from $[2^{15},2^{16}]$, the expected number of primes among them is $20200/10923\approx 18.4931$.Calculating the approximation with the prime number theorem, I get approximately $16.8315$. The logarithms are supposed to be base-$e$: if I redo the calculation with base-$10$ logarithms, I get approximately $38.7559$.