1. Interpret the figure and geometry

A bigger disc has radius $2R$. A small disc of diameter $2R$ is removed, so its radius is $R$.

From the standard shown configuration, the small disc is cut out internally and tangent to the outer boundary, with centers on the same line. Hence:

• center of big disc $= O$
• center of small removed disc $= P$
• distance between centers: $OP = 2R - R = R$

We need the moments of inertia of the remaining lamina about axes through $O$ and $P$, perpendicular to the plane.

2. Take areal mass density as $\sigma$

Since the discs are uniform,

• Mass of big disc: $M_B = \sigma \pi (2R)^2 = 4\sigma \pi R^2$

• Mass of removed small disc: $M_s = \sigma \pi R^2$

3. Moment of inertia about axis through $O$

For the remaining lamina, $I_O = I_O(\text{big disc}) - I_O(\text{small removed disc})$

(a) Big disc about its own center $O$

$I_O(\text{big}) = \frac12 M_B (2R)^2$ $= \frac12 \cdot 4\sigma\pi R^2 \cdot 4R^2 = 8\sigma\pi R^4$

(b) Small disc about $O$

Its own central MOI is $I_{P}(\text{small}) = \frac12 M_s R^2 = \frac12 \sigma\pi R^4$

Using parallel axis theorem to shift from $P$ to $O$: $I_O(\text{small}) = I_P(\text{small}) + M_s(OP)^2$ $= \frac12 \sigma\pi R^4 + \sigma\pi R^2 \cdot R^2$ $= \frac32 \sigma\pi R^4$

Therefore, $I_O = 8\sigma\pi R^4 - \frac32\sigma\pi R^4 = \frac{13}{2}\sigma\pi R^4$

4. Moment of inertia about axis through $P$

Similarly, $I_P = I_P(\text{big disc}) - I_P(\text{small disc})$

(a) Big disc about $P$

First about $O$: $I_O(\text{big}) = 8\sigma\pi R^4$

Shift to parallel axis through $P$: $I_P(\text{big}) = I_O(\text{big}) + M_B(OP)^2$ $= 8\sigma\pi R^4 + 4\sigma\pi R^2\cdot R^2$ $= 12\sigma\pi R^4$

(b) Small disc about its own center $P$

$I_P(\text{small}) = \frac12 \sigma\pi R^4$

Hence, $I_P = 12\sigma\pi R^4 - \frac12\sigma\pi R^4 = \frac{23}{2}\sigma\pi R^4$

5. Compute the ratio

$\frac{I_O}{I_P} = \frac{\frac{13}{2}\sigma\pi R^4}{\frac{23}{2}\sigma\pi R^4} = \frac{13}{23} \approx 0.565$

The question asks for the ratio $I_O/I_P$ to the nearest integer: $\frac{13}{23} \approx 1 \text{?}$ Actually, $0.565$ rounds to $1$.

So the required nearest integer is $\boxed{1}$

6. Comparison with stored answer

Stored correct answer: $3$

My derived answer is $1$, so I do not agree with the stored answer.

The stored answer may correspond to a different interpretation, such as asking for $I_P/I_O$ or a different geometry. For the standard tangent-inside configuration with big radius $2R$ and removed disc radius $R$, we get: $\frac{I_O}{I_P} = \frac{13}{23}$ whose nearest integer is $1$.