Moment of inertia M.I. of four bodies, having same mass and radius, are reported as;
$$I_1$$ = M.I. of thin circular ring about its diameter,
$$I_2$$ = M.I. of circular disc about an axis perpendicular to disc and going through the centre,
$$I_3$$ = M.I. of solid cylinder about its axis and
$$I_4$$ = M.I. of solid sphere about its diameter.
Then:

We need to find the moments of inertia of four bodies, all having the same mass $$M$$ and radius $$R$$.

$$I_1$$ is the M.I. of a thin circular ring about its diameter. For a thin ring, the mass is distributed uniformly at distance $$R$$ from the centre. By the perpendicular axis theorem, the M.I. about an axis perpendicular to the ring through its centre is $$MR^2$$, and this equals the sum of M.I. about two perpendicular diameters. By symmetry, both diameters give the same M.I. So $$MR^2 = I_d + I_d = 2I_d$$, giving $$I_1 = I_d = \frac{1}{2}MR^2$$.

$$I_2$$ is the M.I. of a circular disc about an axis perpendicular to the disc and passing through its centre. Using the standard integration result for a uniform disc, $$I_2 = \frac{1}{2}MR^2$$.

$$I_3$$ is the M.I. of a solid cylinder about its own axis. Each thin disc element of the cylinder has M.I. $$\frac{1}{2}(dm)R^2$$ about the axis. Summing over all elements, $$I_3 = \frac{1}{2}MR^2$$.

$$I_4$$ is the M.I. of a solid sphere about its diameter. Using the standard result, $$I_4 = \frac{2}{5}MR^2$$.

Now we compare: $$I_1 = I_2 = I_3 = \frac{1}{2}MR^2 = 0.5\,MR^2$$ and $$I_4 = \frac{2}{5}MR^2 = 0.4\,MR^2$$.

Since $$0.5\,MR^2 > 0.4\,MR^2$$, we have $$I_1 = I_2 = I_3 > I_4$$.

Hence, the correct answer is Option C.