= Solution
On $r=c$, the traction $\boldsymbol\sigma\mathbf n$ exerted by the outer fluid on the inner fluid is
$$
\boldsymbol\sigma\mathbf n
=-\frac{3\mu B}{c^3}
(\boldsymbol\Omega\times\mathbf n).
$$
The inner fluid exerts the opposite traction on the outer fluid. Its couple is therefore
$$
\begin{aligned}
\mathbf G
&=-\int_{r=c}\mathbf x\times
(\boldsymbol\sigma\mathbf n)\,dS\\
&=3\mu Bc^{-2}
\int_{r=c}\mathbf n\times
(\boldsymbol\Omega\times\mathbf n)\,dS.
\end{aligned}
$$
Using
$$
\mathbf n\times(\boldsymbol\Omega\times\mathbf n)
=\boldsymbol\Omega-\mathbf n(\mathbf n\cdot\boldsymbol\Omega)
$$
and the supplied spherical integral,
$$
\int_{r=c}n_in_j\,dS
=\frac{4\pi c^2}{3}\delta_{ij},
$$
we obtain
$$
\int_{r=c}\mathbf n\times
(\boldsymbol\Omega\times\mathbf n)\,dS
=\frac{8\pi c^2}{3}\boldsymbol\Omega.
$$
Thus the radius cancels:
$$
\mathbf G=8\pi\mu B\boldsymbol\Omega
=\boxed{
\frac{8\pi\mu a^3b^3}{b^3-a^3}
\boldsymbol\Omega}.
$$
This is the <Torque in rotational Stokes flow between concentric spheres>.
When $a\ll b$,
$$
\mathbf G\sim8\pi\mu a^3\boldsymbol\Omega,
$$
the rotational drag torque for a sphere in an effectively unbounded fluid. If $h=b-a\ll a$, then
$$
b^3-a^3\sim3a^2h,
\qquad
a^3b^3\sim a^6,
$$
so
$$
\mathbf G\sim
\frac{8\pi\mu a^4}{3h}\boldsymbol\Omega,
$$
the inverse-gap growth expected from a thin <Couette flow>.
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