The nonstoichiometric compound ${\mathrm{Fe}}_{0.94}\mathrm{O}$ is formed when $\mathrm{X}\%\text{ of }{\mathrm{Fe}}^{+2}$ions are replaced by as many as $2/3{\mathrm{Fe}}^{+3}$ ions, then X is 

$$\begin{gathered} \mathrm{X}\% \mathrm{of} {\mathrm{Fe}}^{2+} \mathrm{ions} \mathrm{are} \mathrm{replaced} \mathrm{by} \mathrm{as} \mathrm{many} \frac{2}{3} {\mathrm{Fe}}^{3+} \mathrm{ions} \\ \mathrm{The} \mathrm{number} \mathrm{of} {\mathrm{Fe}}^{3+} \mathrm{replacing} {\mathrm{XFe}}^{2+} \mathrm{ions}=\frac{2\mathrm{x}}{3} \\ \mathrm{The} \mathrm{vacancies} \mathrm{of} \mathrm{cations}=\mathrm{x}-\frac{2\mathrm{x}}{3}=\frac{\mathrm{x}}{3} \\ \mathrm{But} \mathrm{the} \mathrm{non}-\mathrm{stoichiometric} \mathrm{compound} \mathrm{has} \mathrm{formula} {\mathrm{Fe}}_{0.94}\mathrm{O} \\ \frac{\mathrm{x}}{3}=1-0.94=0.06 \\ \mathrm{x}=0.06\times 3=0.18=18\% \end{gathered}$$