Ignore, Wrong Solution.

http://www.komal.hu/verseny/feladat.cgi?a=feladat&f=B4582&l=en

I think that if $n^3 = \prod_{i=1}^{k} p_i^{3\alpha_i}$, thus $d(n^3) = \prod_{i=1}^{k}(3\alpha_i+1)$. Therefore, you must solve $\prod_{i=1}^{k} (3\alpha_i + 1) = 5 \prod_{i=1}^k \alpha_i$. It is not hard to see that $k \le 2$. Indeed, if $k = 1$, it is easy to see that there are no solutions. If $k = 2$, the only solutions are of the form $n = p^3q$.