= Solution
Put $r_S=2R_g$. On the midplane, <circular orbit> balance gives $r\Omega^2=\partial_r\Phi=GM/(r-r_S)^2$. The <specific angular momentum>, <specific orbital energy> and <orbital shear parameter> of a <Paczyński-Wiita circular orbit> are therefore
$$
\boxed{\Omega^2=\frac{GM}{r(r-2R_g)^2}},\qquad
\boxed{h=r^2\Omega=\frac{\sqrt{GM}\,r^{3/2}}{r-2R_g}},
$$
$$
\boxed{\varepsilon=\frac12r^2\Omega^2+\Phi(r,0)
=\frac{GM(4R_g-r)}{2(r-2R_g)^2}},\qquad
\boxed{q=-\frac{d\log\Omega}{d\log r}
=\frac12+\frac r{r-2R_g}}.
$$
For a small radial displacement at fixed <specific angular momentum>, the <effective potential stability criterion> gives the <radial epicyclic frequency>
$$
\boxed{\kappa_r^2=\frac1{r^3}\frac{d(h^2)}{dr}
=\frac{GM(r-6R_g)}{r(r-2R_g)^3}
=2(2-q)\Omega^2}.
$$
Here $\kappa_r$ is the question's epicyclic $\kappa$. Its square is negative for $2R_g<r<6R_g$, so a radial displacement grows instead of undergoing <epicyclic motion>. The orbit at $r_{\rm in}=6R_g$ is marginal, and the outer orbits are stable to small radial displacements. At this inner edge,
$$
\boxed{\varepsilon(6R_g)=-\frac{GM}{16R_g}=-\frac{c^2}{16}},\qquad
\boxed{\eta=\frac1{16}}.
$$
This efficiency belongs to the <Paczyński-Wiita potential>; it is an approximation to the relativistic black-hole result.
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