The moment of inertia about an axis through A and perpendicular to the page is

$\begin{align*}I_A = \left(2/3\sqrt{3/4}l\right)^2 m^2 \cdot 3 = m l^2\end{align*}$
The moment of inertia about an axis through B and perpendicular to the page is
$\begin{align*}I_B = 2 l^2 m\end{align*}$
The kinetic energy is given by
$\begin{align*}K = \frac{1}{2} I \omega^2\end{align*}$
So, the desired ratio is
$\begin{align*}\frac{K_B}{K_A} = \frac{\displaystyle\frac{1}{2} I_B \omega^2}{\displaystyle \frac{1}{2} I_A \omega^2} = \frac{...$
Therefore, answer (B) is correct.

Solution to 1996 Problem 32