= Solution
Write the supplied <velocity> in the radial form
$$
\mathbf u=f(r)\mathbf U+g(r)(\mathbf U\cdot\mathbf x)\mathbf x,\qquad
f=\frac{3a}{4r}+\frac{a^3}{4r^3},\quad
g=\frac{3a}{4r^3}-\frac{3a^3}{4r^5}.
$$
At $r=a$, $f=1$ and $g=0$, giving the <no-slip boundary condition> $\mathbf u=\mathbf U$. At infinity $\mathbf u\to0$ and the supplied gauge-fixed <pressure> tends to zero. These are the laboratory-frame conditions for a sphere translating through otherwise quiescent fluid.
The exterior <Stokes equations> can also be checked directly:
$$
\nabla\cdot\mathbf u=(\mathbf U\cdot\mathbf x)\left(\frac{f'}r+rg'+4g\right)=0,
$$
and
$$
\begin{aligned}
\nabla^2\mathbf u
&=\left(f''+\frac{2f'}r+2g\right)\mathbf U
+\left(g''+\frac{6g'}r\right)(\mathbf U\cdot\mathbf x)\mathbf x\\
&=\frac{3a}{2}\left[\frac{\mathbf U}{r^3}
-\frac{3(\mathbf U\cdot\mathbf x)\mathbf x}{r^5}\right]
=\frac1\mu\nabla p.
\end{aligned}
$$
To compute <surface traction in translating-sphere Stokes flow>, let $\mathbf n=\mathbf x/a$ point from the sphere into the fluid. At the surface, $f'(a)=-3/(2a)$, $g'(a)=3/(2a^3)$ and $g(a)=0$. Hence
$$
\left.\partial_j u_i\right|_{a}
=\frac{3}{2a}\left[-U_i n_j+(\mathbf U\cdot\mathbf n)n_i n_j\right],\qquad
p(a)=\frac{3\mu}{2a}(\mathbf U\cdot\mathbf n).
$$
Substitute into the <Newtonian fluid stress tensor>. Its viscous contribution to the surface <traction> is $3\mu[-\mathbf U+(\mathbf U\cdot\mathbf n)\mathbf n]/(2a)$, so the normal part cancels the <pressure> contribution:
$$
\boxed{\boldsymbol\sigma\mathbf n=-\frac{3\mu}{2a}\mathbf U.}
$$
The vector is constant over the entire sphere in the stated pressure gauge. Integrating it gives the fluid's <drag force> on the sphere,
$$
\boxed{\mathbf F_H=\int_{r=a}\boldsymbol\sigma\mathbf n\,dS
=-\frac{3\mu}{2a}(4\pi a^2)\mathbf U=-6\pi\mu a\mathbf U.}
$$
This is <Stokes's law>. On an exterior fluid domain the outward domain normal at its inner boundary is $-\mathbf n$; that distinction will matter in the reciprocal calculation.
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