Explanation

(c) According to problem disc is melted and recasted into a solid sphere so their volume will be same
$V_{Disc} = V_{Sphere} \Rightarrow \pi R_{Disc}^{2}t = \frac{4}{3}\pi R_{Sphere}^{3}$
$\Rightarrow \pi R_{Disc}^{2}\left( \frac{R_{Disc}}{6} \right) = \frac{4}{3}\pi R_{Sphere\ }^{3}\ \ \ \left\lbrack t = \frac{R_{Disc}}{6},\ given \right\rbrack$
$\Rightarrow R_{Disc}^{3} = 8R_{Sphere\ }^{3} \Rightarrow R_{Sphere} = \frac{R_{Disc}}{2}$
Moment of inertia of disc $I_{Disc} = \frac{1}{2}MR_{Disc}^{2} = I$ [Given]
$\therefore M\left( R_{Disc} \right)^{2} = 2I$
Moment of inertia of sphere $I_{Sphere} = \frac{2}{5}MR_{Sphere}^{2}$
$= \frac{2}{5}M\left( \frac{R_{Disc}}{2} \right)^{2} = \frac{M}{10}\left( R_{Disc} \right)^{2} = \frac{2I}{10} = \frac{I}{5}$