= Solution
There are four cation sites per conventional <unit cell>, hence $0.4$ Gd substitutions per cell on average. Two substitutions of Gd$^{3+}$ for Ce$^{4+}$ require one <oxygen vacancy> for <charge neutrality>, so the cell contains $0.2$ vacancies. Equivalently, the composition is $\mathrm{Ce_{0.9}Gd_{0.1}O_{1.95}}$. Therefore
$$
\boxed{n_v=\frac{0.2}{a^3}=3.461\times10^{27}\,\mathrm{m^{-3}}}.
$$
This uses the supplied lattice parameter literally and treats it as the conventional fluorite-cell edge.
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