Circle: $x^2+y^2-4x+6y-12=0$ has centre $(2,-3)$ and radius $\sqrt{4+9+12}=5$.

A diameter of this circle is a chord of $S$. The diameter passes through $(2,-3)$ with length $10$, so the chord of $S$ has length $10$, meaning half-chord $=5$.

Centre of $S$ is $(-3,2)$. Distance from $(-3,2)$ to centre $(2,-3)$: $d=\sqrt{25+25}=5\sqrt{2}$.

The chord (diameter of inner circle) has its midpoint at $(2,-3)$, and $S$'s centre is at $(-3,2)$. Radius of $S$: $R=\sqrt{d^2+5^2}$... but wait: the diameter of the inner circle passes through $(2,-3)$. Distance from $(-3,2)$ to $(2,-3)$ is $5\sqrt{2}$. The chord (full diameter $=10$) is a chord of $S$, so $R^2=(5)^2+(5\sqrt{2})^2=25+50=75$, $R=5\sqrt{3}$.