From Biot-Savart's law, magnetic field due to current carrying conductor is

$B= \frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{i}{r} \left(\sin {\mathrm{\varphi }}_1+\sin {\mathrm{\varphi }}_2\right)$

Where, ${\mathrm{\varphi }}_1=45°,{\mathrm{\varphi }}_2=45°$

$B=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }} .\frac{i}{{\displaystyle \raisebox{1ex}{$L$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)=\frac{{\mathrm{\mu }}_{0}i\sqrt{2}}{2\mathrm{\pi }L}$

Since there are four sides, the total magnetic field is

$B_{\mathrm{net}}=4B=\frac{2\sqrt{2}{\mathrm{\mu }}_{0}i}{\mathrm{\pi }L}$