First, we find that moment of inertia of the hoop about the pivot point using the parallel axis theorem
$\begin{align*}I = I_{CM} + M h^2 = M (0.2 \mbox{ m})^2 + M (0.2 \mbox{ m})^2 = 2 \cdot 3 \mbox{ kg} \cdot 0.2 \mbox{ m})^2 = ...$
So, the period is
$\begin{align*}2 \pi \sqrt{I/m g d} = 2 \pi \sqrt{0.24 \;\mathrm{kg}\;\mathrm{m}^2/(3 \mbox{ kg} \cdot 9.8 \;\mathrm{m}\mathrm...$
Therefore, answer (C) is correct.

Solution to 1996 Problem 21