1. Moment of inertia of the original solid cylinder

For a solid cylinder about its own axis, $I = \frac{1}{2}MR^2$

So for the original cylinder, $I_1 = \frac{1}{2} M_1 R^2$ where $M_1$ is the mass of the original cylinder.

2. Mass of the carved out cylinder

The carved out portion is also a solid cylinder, concentric with the original one, with $R' = \frac{R}{2}, \qquad L' = \frac{L}{2}$

Since the material is uniform, mass is proportional to volume: $M \propto \pi R^2 L$

Hence, $\frac{M_2}{M_1} = \frac{\pi (R')^2 L'}{\pi R^2 L}$

Substitute $R' = \frac{R}{2}$ and $L' = \frac{L}{2}$: $\frac{M_2}{M_1} = \frac{\left(\frac{R}{2}\right)^2 \cdot \frac{L}{2}}{R^2 L}$ $= \frac{\frac{R^2}{4} \cdot \frac{L}{2}}{R^2 L} = \frac{1}{8}$

Thus, $M_2 = \frac{M_1}{8}$

3. Moment of inertia of the carved out portion

Again using $I = \frac{1}{2}MR^2$ for the carved cylinder, $I_2 = \frac{1}{2} M_2 (R')^2$

Substitute $M_2 = \frac{M_1}{8}$ and $R' = \frac{R}{2}$: $I_2 = \frac{1}{2} \cdot \frac{M_1}{8} \cdot \left(\frac{R}{2}\right)^2$ $= \frac{1}{2} \cdot \frac{M_1}{8} \cdot \frac{R^2}{4}$ $= \frac{M_1 R^2}{64}$

4. Find the ratio $\frac{I_1}{I_2}$

We have $I_1 = \frac{1}{2} M_1 R^2$ and $I_2 = \frac{M_1 R^2}{64}$

Therefore, $\frac{I_1}{I_2} = \frac{\frac{1}{2} M_1 R^2}{\frac{M_1 R^2}{64}} = \frac{1}{2} \times 64 = 32$

5. Final answer

$\boxed{32}$

The derived answer matches the stored correct answer.