A uniform solid cylinder with radius R and length L has moment of...

Moment of inertia of solid cylinder of radius $R$ and length $L$ about the axis is ${I}_{1} = \dfrac{{M}_{1}{R}^{2}}{2}$.

Moment of inertia of carved out portion of the cylinder of radius $\dfrac{R}{2}$ and length $\dfrac{L}{2}$ is ${I}_{2} = \dfrac{{M}_{2}{\left(\dfrac{R}{2}\right)}^{2}}{2} = \dfrac{{M}_{2}{R}^{2}}{8}$

Now, the ratio $\dfrac{{I}_{1}}{{I}_{2}} = \dfrac{4{M}_{1}}{{M}_{2}}$

As, $mass = density \times volume$

So, $\dfrac{{I}_{1}}{{I}_{2}} = \dfrac{4 \times ρ\pi{R}^{2}L}{\dfrac{ρ\pi{R}^{2}L}{4 \times 2}} = 32$