= Solution
Write $Q=U_jx_j$ and use repeated-index summation. Direct differentiation gives
$$
\partial_ju_i=\frac a2\left[-\frac{U_ix_j}{r^3}+\frac{U_jx_i}{r^3}
+\frac{Q\delta_{ij}}{r^3}-\frac{3Qx_ix_j}{r^5}\right].
$$
Tracing yields $\partial_i u_i=0$. Since $\nabla^2(1/r)=0$ for $r>a$, and $\nabla^2(Qx_i/r^3)=2U_i/r^3-6Qx_i/r^5$, we get
$$
\mu\nabla^2u_i=\mu a\left(\frac{U_i}{r^3}-\frac{3Qx_i}{r^5}\right)=\partial_i p.
$$
Hence the incompressible <Stokes equations> hold everywhere outside the bubble.
The Newtonian <viscous stress tensor> $T_{ij}=-p\delta_{ij}+\mu(\partial_ju_i+\partial_iu_j)$ simplifies to
$$
\boxed{T_{ij}=-\frac{3\mu aQx_ix_j}{r^5}.}
$$
At $r=a$ with $n=x/a$, $u=\tfrac12[U+(U\cdot n)n]$, so $u\cdot n=U\cdot n$: the interface is impermeable in the bubble frame. Also $Tn=-(3\mu/a)(U\cdot n)n$ is purely normal, so the tangential traction vanishes, appropriate to a clean bubble with negligible internal viscosity. The tangential fluid <velocity> need not equal the bubble's <velocity>. At infinity $u,p\to0$. The calculation assumes the usual spherical, negligible-deformation <limit>; exact normal-stress balance includes interior <pressure>, <surface tension> and any buoyancy sustaining the motion.
Integrate the outer-fluid traction with the normal pointing from the bubble into the fluid:
$$
F_i=-\frac{3\mu}{a}U_j\int n_in_j\,dS=-4\pi\mu aU_i.
$$
Thus $\boxed{F_{\rm drag}=-4\pi\mu aU}$, smaller than the no-slip rigid-<sphere> <Stokes drag law> $-6\pi\mu aU$.
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