Primy Cubes!

$p^3-q^3=(p-q)(p^2+pq+q^2)$

The difference of cubes $p^3-q^3$ is prime only if one of the above factor is $1$ .

$p^2+pq+q^2 > 1$ for any primes.Therefore $p-q$ has to be $1$

$p-q=1$ only for $p=3$ , $q=2$ as they are the only primes that differ by $1$ .

So the required number is $3^3-2^3 = 19$ which indeed is a prime.

Prime numbers are $2,3,5,7$ and so on...... But if you take any 2 odd primes, the difference is an even number, which is a multiple of 2 and has more than 2 factors. So you can only do $3^{3}-2^{3}$ , which is $\boxed{19}$ .