Explanation

(d) Central of the mass of the rod lies at the midpoint and when the is displaced.
Through an angle $60{^\circ}\ $it lises to point B.
From the figure
$\sin{30{^\circ} = \frac{BC}{AB}}$
Or $\ \ \ \ \sin{30{^\circ}} = \frac{1}{l/2}\ $
Or $\frac{1}{2} = \frac{L}{l/2}$
Or $L = \frac{l}{4}$
The potential energy of the rod in this position is
$\ \ \ U = mgL$
$\ \ \ U = mg\frac{1}{4}$