Who can solve this?

As $a,b,c,d,e$ form an arithmetic progression we know that $b + d = a + e = 2c,$ and so

$a + b + c + d + e = 5c = m^{3}$ for some positive integer $m.$

Again, since $b + d = 2c$ we have that $b + c + d = 3c = n^{2}$ for some positive integer $n.$

So $\dfrac{5c}{3c} = \dfrac{m^{3}}{n^{2}} \Longrightarrow 5n^{2} = 3m^{3}.$

To find the minimum value of $c$ we need to keep $m, n$ as small as possible. To this end, we need only have $m,n$ have prime factors $3$ and $5.$ What remains to find is the multiplicities of these factors.

Letting $\large n = 3^{p}5^{q}$ and $\large m = 3^{r}5^{s}$ for non-negative integers $p,q,r,s$ we then have that

$\large 5*(3^{p}5^{q})^{2} = 3(3^{r}5^{s})^{3} \Longrightarrow 3^{2p}5^{2q + 1} = 3^{3r + 1}5^{3s}.$

Equating powers of like exponents, we then require that $2p = 3r + 1$ and $2q + 1 = 3s.$ The least allowed values satisfying these equations are $p = 2, r = 1$ and $q = 1, s = 1,$ giving us $m = 3*5 = 15$ and $n = 3^{2}*5 = 45.$

Then as $5c = m^{3}$ we find that the minimum value for $c$ is $\dfrac{15^{3}}{5} = 3*225 = \boxed{675}.$

(To confirm, with $3c = n^{2}$ we find that $c = \dfrac{n^{2}}{3} = \dfrac{45^{2}}{3} = 15*45 = 675.$ )