From a solid sphere of mass $$M$$ and radius $$R$$, a spherical portion of radius $$\left(\frac{R}{2}\right)$$ is removed as shown in the figure. Taking gravitational potential $$V = 0$$ at $$r = \infty$$, the potential at the centre of the cavity thus formed is ($$G$$ = gravitational constant)

$$V_P = V_1 - V_2$$

The mass of the removed part is  $$m = M\left(\frac{R/2}{R}\right)^3=\frac{M}{8}$$

Potential inside a uniform solid sphere at a distance $$x$$ from its centre is  $$V(x)=-\frac{GM}{2R^3}(3R^2-x^2)$$

Substituting $$x=\frac{R}{2}$$,

$$V_1=-\frac{GM}{2R^3}\left[3R^2-\left(\frac{R}{2}\right)^2\right]$$

$$V_1=-\frac{GM}{2R^3}\left(3R^2-\frac{R^2}{4}\right)$$

$$V_1=-\frac{GM}{2R^3}\left(\frac{11R^2}{4}\right)=-\frac{11GM}{8R}$$

The potential at the centre of a uniform solid sphere of mass $$m$$ and radius $$r$$ is  $$V_2=-\frac{3Gm}{2r}$$

For the removed sphere,  $$m=\frac{M}{8},\qquad r=\frac{R}{2}$$

Therefore,  $$V_2=-\frac{3G(M/8)}{2(R/2)}=-\frac{3GM}{8R}$$

Since the sphere is removed,  $$V_P=V_1-V_2$$

$$V_P=\left(-\frac{11GM}{8R}\right)-\left(-\frac{3GM}{8R}\right)$$

$$V_P=-\frac{GM}{R}$$

Therefore,  $${V_P=-\frac{GM}{R}}$$

Hence, the correct option is C.