Four particles each of mass 'm' are lying symmetrically on the rim of the disc of mass M and radius R moment of inertia of this system about an axis passing through one of the particles and perpendicular to plane of disc is : 

(3M + 16m) (R² / 2)  

axis of rotation is passing through particle (3)

Moment of Inertia of disc about axis = (1/2)MR² + MR² = (3/2)MR² by figure (2),

Moment of Inertia of particle (2) about given axis = m(√2R)² = 2mR² 

Moment of Inertia of particle (4) about given axis = m(√2R)² = 2mR² 

Moment of Inertia of particle (1) about given axis = m(2R)² = 4mR² 

Total moment of inertia = (3/2)MR² + 2mR² + 2mR² + 4mR² 

   = (3/2)MR² + 8mR² 

   = (R²/2) (3M + 16m)