Three rings, each of mass ‘M’ and radius ‘R’ are kept touching each other such that their centres form an equilateral triangle. The M.I. of the system about a median of the triangle is

Ring-1's moment of inertia with regard to the specified axis = $\mathrm{m}\times \frac{{\mathrm{r}}^2}{2}\mathrm{\_\_\_\_\_}\left(1\right)$

where r is the ring's radius and m is the ring's mass.

By the parallel axis theorem, the moment of inertia of ring-2 with respect to the specified axis is = $\mathrm{m}\times \frac{{\mathrm{r}}^2}{2}+\mathrm{m}\times {\mathrm{r}}^2=\frac{3}{2}\mathrm{m}\times {\mathrm{r}}^2\mathrm{\_\_\_\_\_\_}\left(2\right)$

similarly Ring-3's moment of inertia with respect to the specified axis, according to the parallel axis theorem

$=\frac{3}{2}\mathrm{m}\times {\mathrm{r}}^2\mathrm{\_\_\_\_\_\_\_}\left(3\right)$

The system of combined rings' moment of inertia I is given by, I =$\left(\frac{3}{2}+\frac{3}{2}+\frac{1}{2}\right)\mathrm{m}\times {\mathrm{r}}^2=\frac{7}{2}{\mathrm{mr}}^2$

Hence the correct answer $3.5M{R}^2.$