 Solution

For the uniform magnetic field in an off-axis cylindrical cavity, let $\mathbf{b}$ point from the conductor's axis to the hole's axis and let the current flow in $+z$. Its uniform current density in the metal isSuperpose a complete cylinder carrying $\mathbf{J}$ and a cylinder occupying the hole carrying $-\mathbf{J}$. Inside a uniformly filled cylinder, Ampère's law gives $\mathbf{B}=({\mu }_0/2)\mathbf{J}\times \mathbf{r}$. For any point in the cavity,There is no remaining position dependence, so the field is uniform. Its magnitude is ${\mu }_{0}Ib/[2\pi (R^2-a^2)]$, perpendicular to the axis-offset vector in the cross-sectional plane. Reversing the current reverses the field.