MI about a diameter : \\\\(I= \\frac{1}{2}MR^2\\\\) by perpendicular axis theoremMI about an axis passing through tangent to the ring: \\\\(I = \\frac{1}{2}MR^2 + MR^2= \\frac{3}{2}MR^2\\\\) by parallel axis theoremFor maximum radius of gyration ring should have maximum moment of inertia about a axis here, option [D] has maximum moment of inertial which is \\\\(2MR^2\\\\)