According to question , resistance of wire ADC is twice that of wire ABC.

Hence current flowing through ADC

is half that of ABC, i.e., $\frac{{\mathrm{i}}_2}{{\mathrm{i}}_1}=\frac{1}{2}.$  Also i₁ + i₂ = i

$\Rightarrow {\mathrm{i}}_1=\frac{2\mathrm{i}}{3} \mathrm{and} {\mathrm{i}}_2=\frac{\mathrm{i}}{3}$

Magnetic field at centre O due to wire AB and BC (part 1 and 2)

${\mathrm{B}}_1={\mathrm{B}}_2=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}.\frac{2{\mathrm{i}}_1\sin 45\mathrm{°}}{\mathrm{a}/2}\otimes =\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}.\frac{2\sqrt{2}{\mathrm{i}}_1}{\mathrm{a}}\otimes$

and magnetic field at centre O due to wires AD and DC

[i.e. part 3 and 4]  ${\mathrm{B}}_3={\mathrm{B}}_4=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{2\sqrt{2}{\mathrm{i}}_2}{\mathrm{a}}\odot$

Also i₁ = 2i₂ , so ( B₁ = B₂ ) > ( B₃ = B₄ )

Hence net magnetic field at centre O

B_net = ( B₁ + B₂) - ( B₃ + B₄ )

$=2\times \frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}.\frac{2\sqrt{2}\times \left(\frac{2}{3}\mathrm{i}\right)}{\mathrm{a}}-\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}.\frac{2\sqrt{2}\left(\frac{\mathrm{i}}{3}\right)\times 2}{\mathrm{a}}$

$=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}.\frac{4\sqrt{2}\mathrm{i}}{3\mathrm{a}}(2-1)\otimes =\frac{\sqrt{2}{\mathrm{\mu }}_0\mathrm{i}}{3\mathrm{\pi a}}\otimes$