= Solution
Write
$$
f(r)=A+\frac B{r^3},
\qquad
\mathbf w=\boldsymbol\Omega\times\mathbf x,
\qquad
\mathbf u=f\mathbf w.
$$
The velocity gradient of the solid-rotation part is antisymmetric, so it drops out of the <Newtonian fluid stress tensor>. Since
$$
\partial_i f=-\frac{3B}{r^5}x_i,
$$
and $p=0$, the stress is
$$
\boxed{
\sigma_{ij}
=-\frac{3\mu B}{r^5}
\left[
x_i(\boldsymbol\Omega\times\mathbf x)_j
+x_j(\boldsymbol\Omega\times\mathbf x)_i
\right]}.
$$
Equivalently, with $\mathbf n=\mathbf x/r$,
$$
\sigma_{ij}
=-\frac{3\mu B}{r^3}
\left[
n_i(\boldsymbol\Omega\times\mathbf n)_j
+n_j(\boldsymbol\Omega\times\mathbf n)_i
\right].
$$
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