The dissociation of $\text{CaF}_2$ is given by the equilibrium: $\text{CaF}_2 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + 2\text{F}^- (aq)$ Let the molar solubility of $\text{CaF}_2$ in $0.1\text{ M}$ NaF be $S$. The concentration of $\text{Ca}^{2+}$ will be $S$. The concentration of $\text{F}^-$ will be the sum of $\text{F}^-$ from $\text{CaF}_2$ ($2S$) and $\text{F}^-$ from NaF ($0.1\text{ M}$). Due to the common ion effect, the dissociation of $\text{CaF}_2$ is suppressed, making $2S$ negligible compared to $0.1\text{ M}$. Thus, $[\text{F}^-] = 2S + 0.1 \approx 0.1\text{ M}$. The solubility product expression is: $K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2$ Substitute the values: $5.3 \times 10^{-11} = (S)(0.1)^2$ $5.3 \times 10^{-11} = S \times 10^{-2}$ $S = \frac{5.3 \times 10^{-11}}{10^{-2}} = 5.3 \times 10^{-9}\text{ mol L}^{-1}$.