Explanation

(b) Magnetic field at 0 due to
Part (1) : $B_{1} = 0$
Part (2) : $B_{2} = \frac{\mu_{0}}{4\pi}.\frac{\pi\ i}{(a/2)}\bigotimes$ [along $- Z$-axis]
Part (3) : $B_{3} = \frac{\mu_{0}}{4\pi}.\frac{\ i}{(a/2)}( \downarrow )$ [along $- Y$-axis]
Part (4) : $B_{4} = \frac{\mu_{0}}{4\pi}.\frac{\pi i}{(3a/2)}\bigodot$ [along $+ Z$-axis]
Part (5) : $B_{5} = \frac{\mu_{0}}{4\pi}.\frac{\ i}{(3a/2)}( \downarrow )$ [along $- Y$-axis]
$B_{2} - B_{4} = \frac{\mu_{0}}{4\pi}.\frac{\pi i}{a}\left( 2 - \frac{2}{3} \right) = \frac{\mu_{0}i}{3a}\bigotimes$ [along $- Z$-axis]
$B_{3} + B_{5} = \frac{\mu_{0}}{4\pi}.\frac{1}{a}\left( 2 + \frac{2}{3} \right) = \frac{8\mu_{0}i}{12\pi a}( \downarrow )$ [along $- Y$-axis]
Hence net magnetic field
$B_{net} = \sqrt{\left( B_{2} - B_{4} \right)^{2} + \left( B_{3} + B_{5} \right)^{2}}$
$\frac{\mu_{0}i}{3\pi a}\sqrt{\pi^{2} + 4}$