Solubility of $\mathrm{CaF}_2=S$ mole $/ \mathrm{L}$ $$ \begin{aligned} S=\frac{2.34 \times 10^{-3}}{0.1 \times 78} & =\frac{2.34}{78} \times 10^{-2}=3 \times 10^{-4} \mathrm{~mol} / \mathrm{L} \\ K_{S R}\left(C_2 F_2\right)=4 S^3 & =4\left(3 \times 10^{-4}\right)^3 \\ & =108 \times 10^{-12}=0.0108 \times 10^{-8}(\mathrm{~mol} / \mathrm{L})^3 \end{aligned} $$