= Solution
At terminal speed the downward buoyant weight balances the <Stokes drag>:
$$
m_bg=\zeta v_s,\qquad
v_s=\frac{2a^2(\rho_p-\rho_f)g}{9\eta}.
$$
Positive $\rho_p-\rho_f$ gives downward settling; a lighter particle would rise. With $a=10^{-7}\,\mathrm m$, $\rho_p-\rho_f=200\,\mathrm{kg\,m^{-3}}$, $\eta=10^{-3}\,\mathrm{Pa\,s}$ and $g=9.81\,\mathrm{m\,s^{-2}}$,
$$
v_s\simeq4.36\times10^{-9}\,\mathrm{m\,s^{-1}},
$$
about $4.4\,\mathrm{nm\,s^{-1}}$, or $0.38\,\mathrm{mm}$ per day.
The very small Reynolds number makes inertial corrections to <Stokes flow> negligible for an isolated particle. On short particle-scale distances, <Brownian motion> is much faster than this systematic drift. Nevertheless the nonzero settling speed can produce concentration gradients over sufficiently long times and large sample heights; Brownian motion does not cancel gravity.
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