$O D=\frac{l}{2}$

$B=4\left[\frac{\mu_0 i}{4 \pi\left(\frac{l}{2}\right)}\left(\sin 45^{\circ}+\sin 45^{\circ}\right)\right]$

$B=2 \sqrt{2} \frac{\mu_0 i}{\pi !}$

For circle

$2 \pi r=4l$

$r=\frac{2l}{\pi}$

So $B^{\prime}=\frac{\mu_0 i}{2 r}=\frac{\mu_0 i}{2\left(\frac{2 l}{\pi}\right)}$

$\Rightarrow B^{\prime}=\frac{\mu_0 i \pi}{4 I}$

$\Rightarrow$ So $\frac{B}{B^{\prime}}=\frac{8 \sqrt{2}}{\pi^2}$