Find the smallest possible value of the middle number.

Since $a,b,c,d,e$ are consecutive positive integers

$a=n-2,b=n-1,c=n,d=n+1,e=n+2 \quad n\in\mathbb{Z}_{>2}$

We have two conditions

$\begin{cases} b+c+d=p^2\\ a+b+c+d+e=q^3 \end{cases} \qquad p,q\in\mathbb{Z}$

Which translate to

$\begin{cases} 3n=p^2\\ 5n=q^3 \end{cases}$

In the first equation $3n$ is divisible by $3$ , so $p$ is divisible by $3$ too, it follows that $p^2$ is divisible by 9, which implies that $n$ is divisible by $3$ . With a similar reasoning on the second equation, $n$ is divisible by $25$ . We can now write $n=75k \phantom{c} (k\in\mathbb{N})$ , substitute in the original system

$\begin{cases} 225k=p^2\\ 125\cdot 3k=q^3 \end{cases}$

From the first equation $k$ must be a square, from the second $3k$ must be a cube, therefore the smallest such $k$ is $9$

To conclude, the answer is

$c=n=75k=75\cdot 9 = \boxed{675}$