1. Find the side of the largest cube that can be cut from the sphere

For a cube of side $a$ inscribed in a sphere of radius $R$, the space diagonal of the cube equals the diameter of the sphere.

$a\sqrt{3} = 2R$

So,

$a = \frac{2R}{\sqrt{3}}$

2. Find the mass of the cube

Since the cube is cut from a uniform solid sphere, density remains same.

Let density be $\rho$.

Mass of sphere:

$M = \rho \cdot \frac{4}{3}\pi R^3$

Hence,

$\rho = \frac{M}{\frac{4}{3}\pi R^3} = \frac{3M}{4\pi R^3}$

Volume of cube:

$V_{\text{cube}} = a^3 = \left(\frac{2R}{\sqrt{3}}\right)^3 = \frac{8R^3}{3\sqrt{3}}$

Therefore mass of cube is

$m = \rho a^3 = \frac{3M}{4\pi R^3} \cdot \frac{8R^3}{3\sqrt{3}}$

$m = \frac{2M}{\pi\sqrt{3}}$

3. Moment of inertia of the cube about an axis through its center and perpendicular to a face

For a cube (or cuboid) of mass $m$ and side $a$, about an axis through center and perpendicular to one face:

$I = \frac{1}{12}m(a^2+a^2) = \frac{1}{6}ma^2$

Now substitute $m = \frac{2M}{\pi\sqrt{3}}$ and $a = \frac{2R}{\sqrt{3}}$.

First,

$a^2 = \frac{4R^2}{3}$

So,

$I = \frac{1}{6}\cdot \frac{2M}{\pi\sqrt{3}} \cdot \frac{4R^2}{3}$

$I = \frac{8MR^2}{18\pi\sqrt{3}}$

$I = \frac{4MR^2}{9\pi\sqrt{3}}$

Now rewrite the denominator:

$\pi\sqrt{3} = \sqrt{3\pi^2}$

So the simplified standard form is

$I = \frac{4MR^2}{9\pi\sqrt{3}}$

This does not match any option exactly as written. But option A is likely intended as

$\frac{4MR^2}{9\sqrt{3}\,\pi}$

which is algebraically the same as our result.

Thus the correct option is A.

4. Option check

• A: $\displaystyle \frac{4MR^2}{9\sqrt{3}\,\pi}$ if interpreted properly, matches.
• B: four times larger than correct value.
• C, D: clearly inconsistent in dimensionless coefficient.

Hence, Option A is correct.