Explanation

(d)
When carbon is bonded to four other atoms, the angle between any pair of bonds $= 109{^\circ},28'$ (tetrahedral angle) but the ring of cyclobutane is square with four angles of $90{^\circ}$.
So, deviation of the bond angle (angle strain) in cyclobutane
$= 109{^\circ}28^{'} - \frac{90{^\circ}}{2}$
$= \frac{19{^\circ}28'}{2}$
$= 9{^\circ}44'$