= Solution
The <electric polarization> $\mathbf P$ is electric dipole moment per unit volume. It produces
$$
\rho_{\rm b}=-\nabla\cdot\mathbf P,
\qquad
\sigma_{\rm b}=\mathbf P\cdot\mathbf n.
$$
The <electric displacement field> $\mathbf D=\epsilon_0\mathbf E+\mathbf P$ is useful because
$$
\nabla\cdot\mathbf D=\rho_{\rm free},
$$
so it separates free charge from the bound charge represented by $\mathbf P$. In a linear isotropic dielectric, $\mathbf D=\epsilon\mathbf E$.
Spherical symmetry and Gauss's law for $\mathbf D$ give, on both sides of the interface,
$$
\mathbf D=\frac{q}{4\pi r^2}\mathbf e_r.
$$
Thus
$$
\mathbf P_i
=\left(1-\frac{\epsilon_0}{\epsilon_i}\right)
\frac{q}{4\pi r^2}\mathbf e_r.
$$
With $\mathbf n=\mathbf e_r$ pointing from medium one to medium two, the net bound sheet charge at $r=R$ is the polarization jump
$$
\boxed{
\sigma_{\rm b}
=(\mathbf P_1-\mathbf P_2)\cdot\mathbf n
=\frac{\epsilon_0q}{4\pi R^2}
\left(\frac1{\epsilon_2}-\frac1{\epsilon_1}\right)}.
$$
If one assigns surface charge separately to the inner body and outer medium, their contributions are $\mathbf P_1\cdot\mathbf n$ and $-\mathbf P_2\cdot\mathbf n$; their sum is the displayed physical interface charge.
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