A long straight metal rod has a very long hole of radius ‘a ’ drilled parallel to the rod  carries a current i . Find the magnetic field on axis of hole. Given C  is the center of the hole and  OC=c

In the rod current density.

$j=\frac{i}{\text{π}(b^2-a^2)}$

Actual field is the vector sum of two current carrying rods, having same current density but in opposite direction. On the hole axis, only the larger rod contributes magnetic field.
Imagine an ampere a loop of radius E and apply Ampere law.
$\begin{array}{l}\oint B.\text{d}l={\mu }_0{i}_{\text{enclosed}}\text{}\\ B2\text{π}c={\mu }_0\left(j\text{π}{c}^2\right)\text{}\\ or B=\frac{{\mu }_{0}i\text{π}{c}^2}{2\text{π}c\text{π}\left(b^2-a^2\right)}=\frac{{\mu }_{0}ic}{2\text{π}\left(b^2-a^2\right)}\text{}\\ or B=\frac{{\mu }_{0}ic}{2\text{π}\left(b^2-a^2\right)}\text{}\\ Hence option \left(a\right) is correct.\end{array}$