= Solution
First convert the <dynamic viscosity> to SI units:
$$
\eta=0.1\frac{10^{-3}\ \mathrm{kg}}{10^{-2}\ \mathrm m\,\mathrm s}=0.01\ \mathrm{Pa\,s}.
$$
Using the stipulated effective-size version of <Stokes's law> with $a=20\times10^{-6}\ \mathrm m$, the drag coefficient is
$$
\zeta=6\pi\eta a=6\pi(0.01)(20\times10^{-6})\approx3.76991\times10^{-6}\ \mathrm{N\,s\,m^{-1}}.
$$
The speed is $v=10^{-7}\ \mathrm{m\,s^{-1}}$, hence
$$
\boxed{F_{\mathrm{drag}}=\zeta v\approx3.76991\times10^{-13}\ \mathrm N=3.76991\times10^{-4}\ \mathrm{nN}}.
$$
The drag acts opposite the velocity. Its magnitude is about $1.06\times10^5$ times smaller than the substrate force in (a); balancing surrounding-fluid drag alone therefore cannot account for the much larger substrate <traction>. The calculation uses the size parameter exactly as supplied. An exact drag calculation for a cube moving close to a wall would need geometry and boundary corrections beyond this stipulated approximation.
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