Answer
$V=\frac{S\sqrt S}{6\sqrt \pi}$
Work Step by Step
Given: $V=3^{-1}(4\pi)^{-1/2}(S^3)^{1/2}\\=\frac{1}{3}.\frac{1}{(4\pi)^{1/2}}.(S^{3\times\frac{1}{2}})\\=\frac{1}{3}.\frac{1}{(4\pi)^{1/2}}.(S^{\frac{3}{2}})\\=\frac{1}{3}.\frac{1}{\sqrt 4\pi}(\sqrt S^3)\\=\frac{1}{3}\frac{1}{2\sqrt \pi}(S\sqrt S)\\=\frac{S\sqrt S}{6\sqrt \pi}$
