From a circular disc of radius R and mass 9M, a small disc of mass M and radius R/3 is removed concentrically. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its centre is

Mass of the disc = 9M
Mass of removed portion of disc = M
The moment of inertia of the complete disc about an axis passing through its centre O and perpendicular to its plane is
I1 = (9/2)MR2
Now, the moment of inertia of the disc with removed portion
I2 = (1/2)M(R/3)2 = (1/18)MR2
Therefore, moment of inertia of the remaining portion of disc about O is
I = I1 - I2 = (9MR2)/2 - (MR2)/18 = (40MR2)/9