detailed solution

Correct option is A

As the point O is on the line AD, the magnetic field at O due to AD is zero. Similarly, the field at O due to BC is also zero. The field at the centre of a circular current loop is given  by $\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{i}}{2\mathrm{a}}$ . The field due to the circular arc BA will be ${\mathrm{B}}_1=\left(\frac{\mathrm{\theta }}{2\mathrm{\pi }}\right)\left(\frac{{\mathrm{\mu }}_0\mathrm{i}}{2\mathrm{a}}\right)$.
The right-hand thumb rule shows that the field is coming out of the plane of the figure. The field due to the circular arc DC is ${\mathrm{B}}_2=\left(\frac{\mathrm{\theta }}{2\mathrm{\pi }}\right)\left(\frac{{\mathrm{\mu }}_0\mathrm{i}}{2\mathrm{b}}\right)$ .
Going into the plane of the figure. The resultant field at O is $\mathrm{B}={\mathrm{B}}_1-{\mathrm{B}}_2=\frac{{\mathrm{\mu }}_0\mathrm{i\theta }(\mathrm{b}-\mathrm{a})}{4\mathrm{\pi ab}}$
coming out of the plane.