Moment of inertia of a solid cylinder of mass $M$, length $2 R$ and radius $R$ about an axis passing through the centre of mass and perpendicular to the axis is
$$
\begin{aligned}
& I_1=M\left(\frac{L^2}{12}+\frac{R^2}{4}\right) \\
& I_1=M\left[\frac{4 R^2}{12}+\frac{R^2}{4}\right]=M\left[\frac{7 R^2}{12}\right]
\end{aligned}
$$
Moment of inertia passing through one end of the cylinder and perpendicular to the axis of the cylinder
$$
\begin{aligned}
I_2 & =M\left(\frac{L^2}{3}+\frac{R^2}{4}\right) \\
I_2 & =M\left[\frac{4 R^2}{3}+\frac{R^2}{4}\right] \\
& =M\left[\frac{19 R^2}{12}\right] \\
\therefore \quad I_2-I_1 & =M R^2
\end{aligned}
$$