= 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 is
$$
\mathbf J=\frac{I}{\pi(R^2-a^2)}\hat{\mathbf z}.
$$
Superpose 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,
$$
\mathbf B=\frac{\mu_0}{2}\bigl[\mathbf J\times\mathbf r-\mathbf J\times(\mathbf r-\mathbf b)\bigr]
=\boxed{\frac{\mu_0I}{2\pi(R^2-a^2)}\,\hat{\mathbf z}\times\mathbf b.}
$$
There is no remaining position dependence, so the field is uniform. Its magnitude is $\mu_0Ib/[2\pi(R^2-a^2)]$, perpendicular to the axis-offset <vector> in the cross-sectional plane. Reversing the current reverses the field.
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