Let $r$ be the radius of the base, $h$ the vertical height so that $\sqrt{r^2+h^2} $ is the slant height. Then $S=\pi r\sqrt{r^2+h^2}$ and $V=\frac{1}{3}\pi r^2h$. Then \[ \left(\frac{6V}{\pi}\right)^{2}\leq\left(\frac{2S}{\pi\sqrt 3}\right)^{3}\] \[ \iff (2r^h)^{2}\le \left(\frac{2r\sqrt{r^2+h^2}}{\sqrt 3}\right)^{3}\] \[ \iff 3\sqrt{3}rh^2 \le 2(\sqrt{r^2+h^2})^3\] Squaring, this becomes \[ 27r^2h^4 \le 4(r^2+h^2)^3\] Now let $r^2=a$ and $h^2=b$. Then after expanding the inequality becomes \[15ab^2 \le 4a^3+12a^2b+4b^3 \] Which is true since $(2a-b)^2(a+4b)\ge 0$. Hence equality occurs when $2a=b$, i.e. $2r^2=h^2$. In this case the slant height is equal to $\sqrt{3}r$.