The solubility product of calcium fluoride, $\ce{CaF2},$ is $\pu{1.46E-10 mol^3 dm^{-9}.}$
What mass of calcium fluoride will dissolve in $\pu{500 cm^3}$ of $\pu{0.10 mol dm^{-3}}$ sodium fluoride solution? (Molar mass of $\ce{CaF2}$ is $\pu{78.1 g mol^{-1}}.)$
I tried using the following method:
$$ \begin{align} [\ce{Ca^{2+}}][\ce{F^-}]^2 &= K_\mathrm{sp}(\ce{CaF2})\\ x(2x)^2 &= \pu{1.46E-10 mol^3 dm^{-9}} \\ x &= \pu{3.3E-4 mol dm^{-3}} \end{align} $$
Solubility $s$ of $\ce{CaF2}$ can be found as such:
$$s(\ce{CaF2}) = xM(\ce{CaF2}) = \pu{3.3E-4 mol dm^{-3}}\times\pu{78.1 g mol-1} = \pu{2.6E-2 g dm^{-3}}.$$
Finally, the mass of $\ce{CaF2}$ dissolved in $\pu{500 cm^3}$ would be
$$m (\ce{CaF2}) = s(\ce{CaF2})\times V = \pu{2.6E-2 g dm^{-3}}\times\pu{500 cm^3} =\pu{1.3E-2 g}.$$
However, my answer differs from the given answer, that is $\pu{5.7E-7 g}.$
How should I approach this question instead?