M.I. of a thin ring (external diameter D, internal diameter d) about an axis perpendicular to the plane of the ring, is

ANS:B - (D⁴ - d⁴)

The given answer is correct since the MI is about an axis perpendicular to the plane.
Izz=Ixx + Iyy according to the perpendicular axis theorem.
So, we get, Izz= πd^4/64 + πd^4/64 = πd^4/32. I think the guys saying π/64 (D^4-d^4) must be in any confusion, because the same would be the answer if the question would have been said that the axis is along the diameter (xx or yy). But if it is perpendicular then, π/32 (D^4-d^4) ....means option B will be the right answer. Correct me if I'm wrong but with a proper explanation.