Answer:
Option D
Explanation:
First we found moment of inertia (MI) of system using parallel axis theorem about centre of mass, then we use it to find moment of inertia about given axis.
Moment of inertia of an outer disc about the axis through centre is
= MR22+M(2K)2=MR2(4+12)=92MR2

For 6 such discs,
Moment of inertia = 6×92MR2=27MR2
So, moment of inertia of system
= MR22−27MR2=552MR2
Hence,
Ip=552MR2+(7M×9R2)
Ip=1812MR2
Isystem=1812MR2