Equilibrium Constant of a Salt knowing the solubility
The molar solubility $s$ equals:
$$s= \frac{15.9\times 10^{-3}}{78}= 0.2038\times10^{-3} \ce{mol/L}$$

According to the equilibrium: $\ce{CaF2 \leftrightarrows Ca^{2+} + 2F-}$, the  equilibrium constant is:


$$K_{sp}=4s^3$$

By substituting the value of $s$ in the above equation, we find:
$$K_{sp}=0.034\times10^{-9}$$