a
(a)

\(\bar{B}\) due to square part :-

\(\vec{B}\) due to side \(O A\) and \(O C\) will be zero at point \(O\)

\(\bar{B}\) due to side \(A B\) and \(B C\) will be equal so

\(\bar{B}_1=2\left[\frac{\mu_0 i}{4 \pi b}\left(\sin 45^{\circ}+0\right)\right]=\frac{\mu_0 i}{2 \sqrt{2} \pi b} \otimes\)

\(\vec{B}\) due to circular part

\(\bar{B}_2=\frac{\mu_0 i}{2 a}\left[\frac{\left(\frac{3 \pi}{2}\right)}{2 \pi}\right]=\frac{3 \mu_0 i}{8 a} \otimes\)

\(\overline{B_{\text {net }}}=\overline{B_1}+\bar{B}_2=\mu_0 i\left[\frac{3}{8 a}+\frac{1}{2 \sqrt{2} \pi b}\right]=\frac{\mu_0 i}{4 \pi}\left[\frac{3 \pi}{2 a}+\frac{\sqrt{2}}{b}\right]\)