= Solution
The <magnetization> $\mathbf M$ is magnetic dipole moment per unit volume. In magnetostatics it produces
$$
\mathbf J_{\rm b}=\nabla\times\mathbf M,
\qquad
\mathbf K_{\rm b}=\mathbf M\times\mathbf n.
$$
The <magnetic field intensity> $\mathbf H=\mathbf B/\mu_0-\mathbf M$ obeys
$$
\nabla\times\mathbf H=\mathbf J_{\rm free},
$$
so $\mathbf H$ isolates free currents while $\mathbf M$ accounts for bound currents. For a linear isotropic material, $\mathbf B=\mu\mathbf H$.
Ampere's law for a circular loop around the free line current gives
$$
\mathbf H=\frac{I}{2\pi r}\mathbf e_\phi
$$
in both media. Therefore
$$
\mathbf M_i
=\left(\frac{\mu_i}{\mu_0}-1\right)
\frac{I}{2\pi r}\mathbf e_\phi.
$$
At $r=R$, the net bound surface-current density is
$$
\begin{aligned}
\mathbf K_{\rm b}
&=(\mathbf M_1-\mathbf M_2)\times\mathbf e_r\\
&=-\frac{I(\mu_1-\mu_2)}
{2\pi R\mu_0}\mathbf e_z.
\end{aligned}
$$
Hence its magnitude is
$$
\boxed{|\mathbf K_{\rm b}|=
\frac{I|\mu_1-\mu_2|}{2\pi R\mu_0}},
$$
and for a positive current along $+\mathbf e_z$ it points along $-\mathbf e_z$ when $\mu_1>\mu_2$, reversing when $\mu_1<\mu_2$.
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