In the given figure net magnetic field at O will be

 

$B_{net}=\sqrt{{(B_2-B_4)}^2+{(B_3+B_5)}^2}=\frac{{\mu }_{0}i}{3\pi a}\sqrt{{\pi }^2+4}$

Magnetic field at 0 due to 
 Part (1) : $B_1=0$

Part (2): $B_2=\frac{{\mu }_0}{4\pi }.\frac{\pi i}{(a/2)}\otimes$ (along –Z-axis)

Part (3): $B_3=\frac{{\mu }_0}{4\pi }.\frac{i}{(a/2)}(\downarrow )$ (along – Y-axis)

Part (4): $B_4=\frac{{\mu }_0}{4\pi }.\frac{\pi i}{(3a/2)} \odot$ (along +Z-axis)
Part (5): $B_5=\frac{{\mu }_0}{4\pi }.\frac{i}{(3a/2)}(\downarrow )$ (along – Y-axis)

$B_2-B_4=\frac{{\mu }_0}{4\pi }.\frac{\pi i}{a}(2-\frac{2}{3})=\frac{{\mu }_{0}i}{3a}\otimes$ (along – Z-axis)

$B_3+B_5=\frac{{\mu }_0}{4\pi }.\frac{1}{a}(2+\frac{2}{3})=\frac{8{\mu }_{0}i}{12\pi a}(\downarrow )$ (along – Y-axis)

Hence net magnetic field $B_{net}=\sqrt{{(B_2-B_4)}^2+{(B_3+B_5)}^2}=\frac{{\mu }_{0}i}{3\pi a}\sqrt{{\pi }^2+4}$