= Solution
{c}
For the hypothetical <Hubble flow>, <Hubble's law> gives $v=H_0d$, where $d=4\times10^5\,\mathrm{km}$ and $H_0$ is the <Hubble constant>. The accompanying constants sheet is not present in the supplied PDF, so retain its dependence explicitly. Taking $H_0=70\,\mathrm{km\,s^{-1}\,Mpc^{-1}}$ as an illustrative input, and using one million <parsecs> equal to $3.08568\times10^{19}\,\mathrm{km}$,
$$
v=70\frac{4\times10^5}{3.08568\times10^{19}}\,\mathrm{km\,s^{-1}}
=9.074\times10^{-8}\,\mathrm{cm\,s^{-1}}.
$$
A year of $365.25$ days contains $3.15576\times10^7\,\mathrm s$. Therefore
$$
\boxed{v=2.86\left(\frac{H_0}{70\,\mathrm{km\,s^{-1}\,Mpc^{-1}}}\right)
\mathrm{cm\,yr^{-1}}.}
$$
Equivalently, $H_0=100h\,\mathrm{km\,s^{-1}\,Mpc^{-1}}$ gives $v=4.09h\,\mathrm{cm\,yr^{-1}}$. This is a counterfactual recession rate for independently comoving bodies, rather than a prediction for the actual <Earth>-<Moon> orbit.
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