The magnetic induction at point ${}^{\text{'}}\mathrm{O}^{\text{'}}$ in the following fig. will be

Given data:

Formula used:

Detailed Solution:

From the figure we can say that 

$B=B_1+B_2+B_3$

As $B_1=0$

$B_2=\frac{3{\mu }_{0}I}{8r}$

 And 

$B_3=\frac{{\mu }_{0}I}{4\pi r}\left[\sin {\theta }_1+\sin {\theta }_2\right]$

Therefore, 

$B=0+\left[\frac{3}{2}-\frac{1}{\pi }\right]\frac{{\mu }_{0}I}{4R}$

$\Rightarrow B=\left[\frac{3}{2}-\frac{1}{\pi }\right]\frac{{\mu }_{0}I}{4r}$

Therefore, the net magnetic field is $\left[\frac{3}{2}-\frac{1}{\pi }\right]\frac{{\mu }_{0}I}{4r}$.