From a solid sphere of mass $$M$$ and radius $$R$$, a cube of the maximum possible volume is cut. Moment of inertia of cube about an axis passing through its centre and perpendicular to one of its faces is:

The sphere is uniform, so we first write its density in terms of the given mass and radius. The volume of a solid sphere is given by the familiar formula

$$V_{\text{sphere}}=\frac{4}{3}\pi R^{3}.$$

Hence the mass-volume relation gives the density

$$\rho=\frac{M}{V_{\text{sphere}}}=\frac{M}{\dfrac{4}{3}\pi R^{3}} =\frac{3M}{4\pi R^{3}}.$$

Now we fit inside the sphere a cube of the largest possible size. For a cube of side $$a$$ the body diagonal equals $$a\sqrt{3}$$. At the limiting size this diagonal coincides with the diameter of the sphere, so we have

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

Solving for $$a$$ gives

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

We next obtain the mass of the cube. Its volume is

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

Multiplying this volume by the same density $$\rho$$ gives the cube’s mass:

$$m_{\text{cube}}=\rho V_{\text{cube}} =\frac{3M}{4\pi R^{3}}\times\frac{8R^{3}}{3\sqrt{3}} =\frac{2M}{\pi\sqrt{3}}.$$

The question asks for the moment of inertia of the cube about an axis that passes through its centre and is perpendicular to one of its faces (say the axis along the $$z$$-direction through the centre). For a rectangular block of sides $$a$$, $$b$$, $$c$$, the moment of inertia about an axis through its centre perpendicular to the face of side $$a\times b$$ is

$$I=\frac{M(a^{2}+b^{2})}{12}.$$

In a cube, $$a=b=c$$, and for the axis perpendicular to a face we have

$$I_{\text{cube}}=\frac{m_{\text{cube}}(a^{2}+a^{2})}{12} =\frac{m_{\text{cube}}\;2a^{2}}{12} =\frac{m_{\text{cube}}\,a^{2}}{6}.$$

We already know $$m_{\text{cube}}$$ and $$a^{2}$$. First compute $$a^{2}$$:

$$a^{2}=\left(\frac{2R}{\sqrt{3}}\right)^{2}=\frac{4R^{2}}{3}.$$

Substituting $$m_{\text{cube}}=\dfrac{2M}{\pi\sqrt{3}}$$ and $$a^{2}=\dfrac{4R^{2}}{3}$$ into the moment-of-inertia expression, we get

$$I_{\text{cube}}=\frac{1}{6}\left(\frac{2M}{\pi\sqrt{3}}\right) \left(\frac{4R^{2}}{3}\right).$$

Now carry out the multiplication step by step:

$$I_{\text{cube}}=\frac{1}{6}\times\frac{2\times4\,M R^{2}}{\pi\sqrt{3}\times3} =\frac{8M R^{2}}{6\times3\pi\sqrt{3}} =\frac{8M R^{2}}{18\pi\sqrt{3}} =\frac{4M R^{2}}{9\pi\sqrt{3}}.$$

This matches Option D. Hence, the correct answer is Option D.