= Solution
The <magnetization> $M$ is magnetic dipole moment per unit volume. It produces bound volume and <surface current density>
$$
J_{\rm b}=\nabla\times M,
\qquad K_{\rm b}=M\times n.
$$
Starting from the <vector potential> of magnetic dipoles and integrating by parts gives
$$
A(x)=\frac{\mu_0}{4\pi}
\left[
\int_V\frac{\nabla'\times M(x')}{|x-x'|}\,d^3x'
+\int_S\frac{M(x')\times n}{|x-x'|}\,dS'
\right],
$$
which has precisely the form generated by those bound currents.
Write the cylinder radius as $R$ and use cylindrical radius $\rho$. <Ampère's law> for the free axial current gives, inside the material,
$$
H(\rho)=\frac{I}{2\pi\rho}\hat\phi,
\qquad
M=\frac B{\mu_0}-H
=\left(\frac\mu{\mu_0}-1\right)
\frac{I}{2\pi\rho}\hat\phi.
$$
At $\rho=R$ this is tangential to the cylinder. Since $n=\hat r$,
$$
\boxed{K_{\rm b}=M\times\hat r
=-\left(\frac\mu{\mu_0}-1\right)
\frac{I}{2\pi R}\hat z},
$$
and
$$
\boxed{M(R)=\left(\frac\mu{\mu_0}-1\right)
\frac{I}{2\pi R}\hat\phi}.
$$
For $\mu>\mu_0$, the bound surface current therefore runs opposite to the free wire current.
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