= Solution
Each missing oxide ion leaves two units of excess positive effective charge, compensated by reducing two Ce$^{4+}$ ions to Ce$^{3+}$. For $0.2$ <oxygen vacancies> per conventional <unit cell>, $0.4$ of its four cerium ions must be trivalent: \b[10% Ce$^{3+}$]. This assumes the same number of cation sites per unit <volume> as in the doped calculation and that oxygen loss is compensated entirely by this cerium-valence change. If reduction changes the cell <volume> appreciably, matching the absolute vacancy density requires correcting the percentage for that change.
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