${F_1} = \frac{{{\mu _0}}}{{4\pi }}\frac{{2IiL}}{{(L/2)}} = \frac{{{\mu _0}Ii}}{\pi }$

acting towards $X Y$ in the plane of loop. Force on arm $C D$ due to current in conductor $X Y$ is

${F_2} = \frac{{{\mu _0}}}{{4\pi }}\frac{{2IiL}}{{3(L/2)}} = \frac{{{\mu _0}Ii}}{{3\pi }}$

acting away from $X Y$ in the plane of loop.

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

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