= Solution
At terminal speed, the downward excess weight is balanced by <Stokes's law> drag:
$$
\frac{4\pi a^3}{3}(\rho_p-\rho_w)g=6\pi\eta av_t.
$$
The subtraction of the displaced water mass includes <buoyancy>. Hence the <Stokes settling velocity> is
$$
\boxed{v_t=\frac{2(\rho_p-\rho_w)ga^2}{9\eta}}.
$$
A density is needed but is not supplied. For the estimate, take a dense iron oxide represented by <magnetite>, with $\rho_p\simeq5.2\times10^3\,\mathrm{kg\,m^{-3}}$, and water with $\rho_w\simeq10^3\,\mathrm{kg\,m^{-3}}$. This gives
$$
v_t\simeq\frac{2(4.2\times10^3)(9.81)(10^{-6})^2}{9(10^{-3})}
=\boxed{9.2\times10^{-6}\,\mathrm{m\,s^{-1}}\simeq9\,\mu\mathrm{m\,s^{-1}}\quad\text{downwards}}.
$$
Using the diameter for the <particle Reynolds number> gives $\mathrm{Re}\simeq\rho_wv_t(2a)/\eta\simeq1.8\times10^{-5}$, which supports the creeping-flow assumption. The model assumes a dilute isolated sphere far from walls and no imposed flow. The precise numerical speed depends on the unspecified oxide phase and particle density; the estimate is of order $10\,\mu\mathrm{m\,s^{-1}}$.
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