In the given figure net magnetic field at O will be
 

Magnetic field at 0 due to

Part (1) :  $B_1=0$
Part (2): 

$B_2=\frac{{\mu }_0}{4\pi }\cdot \frac{\pi i}{(a/2)}\otimes$

(along –Z-axis)
Part (3):  

$B_3=\frac{{\mu }_0}{4\pi }\cdot \frac{i}{(a/2)}(\downarrow )$

    (along – Y-axis)
Part (4): 

$B_4=\frac{{\mu }_0}{4\pi }\cdot \frac{\pi i}{(3a/2)}\odot$

    (along +Z-axis)
Part (5):   

$B_5=\frac{{\mu }_0}{4\pi }\cdot \frac{i}{(3a/2)}(\downarrow )$

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

 (along – Z-axis)

$B_3+B_5=\frac{{\mu }_0}{4\pi }\cdot \frac{i}{a}\left(2+\frac{2}{3}\right)=\frac{8{\mu }_{0}i}{12\pi a}(\downarrow )$

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