 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 $\mathrm{C}{\mathrm{e}}_{0.9}\mathrm{G}{\mathrm{d}}_{0.1}{\mathrm{O}}_{1.95}$. ThereforeThis uses the supplied lattice parameter literally and treats it as the conventional fluorite-cell edge.