The molar solubility of ${\mathrm{CaF}}_2\left({\mathrm{K}}_{\mathrm{sp}}=5.3\times {10}^{-11}\right)$ in 0.1 M solution of NaF will be

${\mathrm{K}}_{\mathrm{sp}}\left({\mathrm{CaF}}_2\right)=5.3\times {10}^{-11}$

${\mathrm{CaF}}_2\rightleftharpoons {\mathrm{Ca}}^{+2}+{\mathrm{F}}^{-}$

$\begin{array}{l}{\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Ca}}^{2+}\right]{\left[{\mathrm{F}}^{-}\right]}^2=\left(\mathrm{s}\right)(0.1{)}^2\\ (\mathrm{Due} \mathrm{to} \mathrm{common} \mathrm{ion} \mathrm{effect}, [{\mathrm{F}}^{-}] =0.1\mathrm{M})\\ \mathrm{s}=\frac{{\mathrm{K}}_{\mathrm{sp}}}{(0.1{)}^2}=\frac{5.3\times {10}^{-11}}{(0.1{)}^2}=5.3\times {10}^{-9}{\mathrm{molL}}^{-1}\end{array}$