How to calculate mass of calcium fluoride that will dissolve in sodium fluoride solution?
In the $\pu{500 cm^3}$ solution there are is already $\ce{F^-}$ dissolved coming from the $\ce{NaF}$ $\pu{0.10 mol dm^{-3}}$ that was already there. This means that the solubility of $\ce{CaF2}$ is reduced due to the common ion effect. 

NB $\ce{NaF}$ can be considered completely dissociated because it has a very high $K_\mathrm{sp};$ therefore the initial concentration of $\ce{F^-}$ is $\pu{0.10 mol dm^{-3}}.$

$$
\begin{align}
[\ce{Ca^{2+}}][\ce{F^-}]^2 &amp;= K_\mathrm{sp}(\ce{CaF2})\\
x(2x+\color{red}{\pu{0.10 mol dm^{-3}}})^2 &amp;= \pu{1.46E-10 mol^3 dm^{-9}} \\
\end{align}
$$

The issue  here is that to find $x$ you need to solve a cubic equation…

However, $K_\mathrm{sp}$ of $\ce{CaF2}$ is very small and $x$, that is its solubility, must be much less than $\pu{0.10 mol dm^{-3}}!$ This allows you to make the approximation that $2x + \pu{0.10 mol dm^{-3}} ≈ \pu{0.10 mol dm^{-3}}.$ 

$$
\begin{align}
[\ce{Ca^{2+}}][\ce{F^-}]^2 &amp;= K_\mathrm{sp}(\ce{CaF2})\\
x(\color{red}{\pu{0.10 mol dm^{-3}}})^2 &amp;= \pu{1.46E-10 mol^3 dm^{-9}} \\
x &amp;= \pu{1.46E-8 mol dm^{-3}}
\end{align}
$$

Solubility $s$ of $\ce{CaF2}$:

$$s(\ce{CaF2}) = xM(\ce{CaF2}) = \pu{1.46E-8 mol dm^{-3}}\times\pu{78.1 g mol-1} = \pu{1.14E-6 g dm^{-3}}.$$

Finally, the mass of $\ce{CaF2}$ dissolved in $\pu{500 cm^3}$ would be

$$m (\ce{CaF2}) = s(\ce{CaF2})\times V = \pu{1.14E-6 g dm^{-3}}\times\pu{500 cm^3} =\pu{5.7E-7 g}.$$