= Solution
A photon travels at $c$, so its typical scattering interval is
$$
\boxed{
t_{\mathrm{sc}}=\frac{\ell_\gamma}{c}
\simeq\frac{7.5\times10^{10}}{3\times10^8}
\simeq2.5\times10^2\,\mathrm s,
}
$$
about four minutes. The supplied universe age is
$$
t_{\mathrm{age}}=10^4\,\mathrm{yr}
\simeq3.2\times10^{11}\,\mathrm s.
$$
Thus
$$
\boxed{
\frac{t_{\mathrm{sc}}}{t_{\mathrm{age}}}\simeq8\times10^{-10},
\qquad
\frac{t_{\mathrm{age}}}{t_{\mathrm{sc}}}\sim10^9.
}
$$
A photon scatters extremely frequently compared with the expansion timescale, of order the cosmic age. The plasma is opaque and its photons are tightly coupled to the electrons rather than freely streaming over a cosmological distance. In rate language $\Gamma_\gamma=n_e\sigma_ec\gg H$, where $H$ is the <Hubble parameter>.<Cosmology by David Tong>[Section 2, photon decoupling: scattering rate and its comparison with the expansion rate]{ref}
This is a statement about transport and momentum coupling. Elastic <Thomson scattering> alone does not create or destroy photons and hence is not, by itself, proof that an arbitrary photon spectrum reaches complete blackbody <thermal equilibrium>. The interaction interval calculated here is the interval experienced by a photon; the interval per electron differs because there are many more photons than electrons. Even replacing the inconsistent supplied age by the standard age of order a year at this scale factor would leave scattering much faster than expansion.
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