The molar solubility of CaF₂ $\left({\mathrm{K}}_{\mathrm{sp}}=5.3\times {10}^{-11}\right)$in 0.1 M solution of NaF will be

            ${\mathrm{CaF}}_2\rightleftharpoons {\mathrm{Ca}}^{+2}+2{\mathrm{F}}^{-}$
At equili.              S        2S
$\mathrm{NaF}\to {\mathrm{Na}}^{+}+{\mathrm{F}}^{-}$
0.1       0.1      0.1
Due to common ion effect of NaF, solubility of CaF₂ is further suppressed. Therefore the concentration of F^- will be mainly due to NaF
$\begin{array}{l}{\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Ca}}^{+2}\right]{\left[{\mathrm{F}}^{-}\right]}^2\\ {\mathrm{K}}_{\mathrm{sp}}=\left(\mathrm{S}\right)(0.1+2\mathrm{S}{)}^2(\text{ or }\left)\right(0.1+2\mathrm{S}\approx 0.1)\\ \therefore {\mathrm{K}}_{\mathrm{sp}}=\left(\mathrm{S}\right)(0.1{)}^2\end{array}$
$\mathrm{S}=\frac{5.3\times {10}^{-11}}{(0.1{)}^2}=5.3\times {10}^{-9} \mathrm{molelit}{ }^{-1}$