Three identical rings, each of mass M and radius R are placed in the same plane touching each other such that their centers form the vertices of an equilateral triangle. The M.I of the system about an axis passing through center of one of the rings and perpendicular to its plane is

The equation for a ring's moment of inertia about an axis that runs through its centre and perpendicular to its plane is

${\mathrm{I}}_{\mathrm{o}}={\mathrm{MR}}^2\mathrm{\_\_\_\_\_\_}\left(1\right)$

Here, M stands for mass and R for radius.
The moment of inertia of a ring around an axis parallel to the axis Y but separated by d from the axis Y is determined by the parallel axis theorem.

$\mathrm{I}={\mathrm{I}}_{\mathrm{o}}+{\mathrm{Md}}^2$

The system's moment of inertia is,

$\mathrm{I}={\mathrm{I}}_1+{\mathrm{I}}_2+{\mathrm{I}}_3$

$\mathrm{I}=\left({\mathrm{MR}}^2\right)+[{\mathrm{MR}}^2+\mathrm{M}{\left(\mathrm{OA}\right)}^2]+[{\mathrm{MR}}^2+\mathrm{M}{\left(\mathrm{OB}\right)}^2]$

$\mathrm{I}=3{\mathrm{MR}}^2+\mathrm{M}{\left(2\mathrm{R}\right)}^2+\mathrm{M}{\left(2\mathrm{R}\right)}^2 =11{\mathrm{MR}}^2$

Hence the correct answer is $11{\mathrm{MR}}^2.$