A circular disc of radius R and thickness $\frac{R}{6}$ has moment of inertia $l$ about an axis passing through its centre and perpendicular to its plane. It is melted and recasted into a solid sphere. The moment of inertia of the sphere about its diameter as axis of rotation is 

According to problem disc is melted and recasted into a solid sphere so their volume will be same. 
 $$\begin{gathered} V_{Disc}=V_{Sphere}=\pi {R^2}_{Disc}t=\frac{4}{3}\pi {R^3}_{Sphere} \\ {V}_{Disc}=V_{Sphere}=\pi {R^2}_{Disc}\left(\frac{R_{Disc}}{6}\right)=\frac{4}{3}\pi {R^3}_{Sphere} \\ \Rightarrow R_{Disc}^3=8R_{sphere}^3\Rightarrow R_{sphere}=\frac{R_{Disc}}{2} \end{gathered}$$
Moment of inertia of disc  $l_{Disc}=\frac{1}{2}M{R^2}_{Disc}=l$
Moment of inertia of sphere $l_{Sphere}=\frac{2}{5}M{R^2}_{Sphere}$
$l_{Sphere}=\frac{l_{disc}}{5}$