$\underline{\textbf{Solution 1}}$

Let the positive integer mentioned be $a$, so that $a^3 = 16p+1$. Note that $a$ must be odd, because $16p+1$ is odd. Rearrange this expression yields

$$(a-1)(a^2+a+1) = 16p$$

Because $a$ is odd, therefore $(a-1)$ is even and $(a^2+a+1)$ is odd. Then, $(a-1)$ must be some multiple of $16$. However, for $(a-1)$ to be any multiple of $16$ other than $16$ will mean $p$ is not a prime. Therefore, $(a-1) = 16$ and $a = 17$. It follows that

$$p=a^2 + a + 1 = 17^2 + 17 + 1=\boxed{307}$$

$\underline{\textbf{Solution 2}}$

Because $(16p+1)$ is odd, let $16p+1 = (2a+1)^3$. Then

$$16p+1 = (2a+1)^3 = 8a^3+12a^2+6a+1\implies 8p=a(4a^2+6a+3)$$

We know $p$ is a prime number and apparently not an even number. and $(4a^2+6a+3)$ is an odd number, so $a$ must equal $8$. Therefore

$$p=4a^2+6a+3=4\times 8^2+6\times 8+3=\boxed{307}$$