Given, mass of solid cylinder, $M=10 \mathrm{~kg}$
Radius,
$\begin{aligned}
& R=40 \mathrm{~cm}=0.4 \mathrm{~m} \\
& L=9 R=9 \times 0.4=3.6 \mathrm{~m}
\end{aligned}$
Moment of inertia of the solid cylinder about the line passing through its centre and perpendicular to its axis is given as


$\begin{aligned}
I_{\mathrm{COM}} & =M\left[\frac{L^2}{12}+\frac{R^2}{4}\right]=10\left[\frac{(3.6)^2}{12}+\frac{(0.4)^2}{4}\right] \\
& =10[1.08+0.04]=10 \times 1.12 \\
\Rightarrow \quad I_{\mathrm{COM}} & =11.2 \mathrm{~kg}-\mathrm{m}^2
\end{aligned}$
According to parallel axis theorem, moment of inertia of cylinder about a line passing through its edge and perpendicular to its axis is given as
$\begin{aligned}
I^{\prime} & =I_{\mathrm{COM}}+M\left(\frac{L}{2}\right)^2=11.2+10\left(\frac{3.6}{2}\right)^2 \\
& =11.2+10(3.24)=11.2+324=43.6 \mathrm{~kg}-\mathrm{m}^2
\end{aligned}$