$F_{1}=\frac{\mu_{0}}{4 \pi} \frac{2 l i L}{(L / 2)}=\frac{\mu_{0} I i}{\pi}$

acting towards $-17$ in the plane of loop.

Force on arm $CD$ due to current in conductor $XY$ is

$F_{2}=\frac{\mu_{0}}{4 \pi} \frac{2 I i L}{3(L / 2)}=\frac{\mu_{0} I i}{3 \pi}$

acting away from $-ill$ in the plane of loop.

$\therefore$ Net force on the loop $=F_{1}-F_{2}$

$=\frac{\mu_{0} I i}{\pi}\left[1-\frac{1}{3}\right]=\frac{2}{3} \frac{\mu_{0} I i}{\pi}$