Common ion effect math problem
First the question doesn't really make sense. $\ce{CaF_2}$ is virtually insoluble, so a "solution of $\ce{CaF_2}$" is a really odd phrase. 

Skipping this nit-picking, the reactants are 1 L of a $2.0 \times 10^{-5}$ M $\ce{Ca^{2+}}$ and a $0.1$ M $\ce{F^-}$ solution  and $\ce{CaF_2}$ is to be precipitated. 

This is a sloppy use of significant figures since volume of 1L and the molarity of the $\ce{F^-}$ solution are only given to one significant figure. 

Ignoring the significant figure problem too, let's work backwards from the given answer of 946 ml.

For the initial solution we can solve for the moles of $\ce{Ca^{2+}}$ present.  

$\text{moles }\ce{Ca^{2+}} = 1 \rm{L} * 2.0 \times 10^{-5}\frac{moles}{L} = 2.0 \times 10^{-5}\text{ moles}$

Now if we add 946 ml of 0.1 M NaF, then the moles of $\ce{F^-}$ in solution is 0.0946 moles which vastly more than the calcium. So we can assume the precipitated $\ce{CaF2}$ has a negligible effect on the $[\ce{F^-}]$, and calculate the final concentration of $[\ce{F^-}]$ as:

$\ce{[F^-]}_{\text{final}} = \dfrac{0.0946}{1.946} = 0.0486 \text{M}$

Now from the $\rm{K}_{\text{sp}}$ we have

$\ce{[Ca^{2+}]}_{\text{final}} = \dfrac{\rm{K}_{\text{sp}}} {[\ce{F^-}]^2}= \dfrac{1.7 \times 10^{-10}}{0.0486^2} = 7.20 \times 10^{-8}$

$\text{moles }\ce{Ca^{2+}} \text{ final}= (7.20 \times 10^{-8})(1.946) = 1.40 \times 10^{-7}$

The initial moles of $\ce{Ca^{2+}}$ is $2.0 \times 10^{-5}$ which is +/- $0.05 \times 10^{-5}$. Thus so far as I am concerned the solution is a "magic number" and there is no way to get from the problem statement to that value. In other words why not $2.00 \times 10^{-7}$ or $3.00 \times 10^{-8}$ for the final $\ce{Ca^{2+}}$ concentration?!?

Thus you have to assume that you're going to precipitate some fraction of the $\ce{Ca^{2+}}$. So are you going to precipitate 99%, 99.9%, 99.95% or what? There just isn't anything in the problem statement on which you can pin a value.

Just out of curiosity the % $\ce{Ca^{2+}}$ recovered is:

% $\ce{Ca^{2+}}$ recovered = $(1-\dfrac{1.40 \times 10^{-7}}{2\times 10^{-5}}) \times 100\% = 99.3 \%$ 



Note that we can also assume that the original solution is saturated in respect to $\ce{CaF2}$. As shown in OP's post, this leads to a concentration of $2.91 \times 10^{-3} \rm{M}$ for $\ce{F^-}$ and given the 1L volume, $2.91 \times 10^{-3} \rm{moles}$ of $\ce{F^-}$. Adding that to the number of moles from the 946 ml of $\ce{F^-}$ solution gives a slightly different value for the total moles of $\ce{F^-}$:

total moles of $\ce{F^-}$ = $0.0946 + 0.00291 = 0.0975$

$\ce{[F^-]}_{\text{final}} = \dfrac{0.0975}{1.946} = 0.0501 \text{M}$

So the magic value for the textbook solution is to assume that: $\ce{[F^-]}_{\text{final}}= 0.0500$

$\ce{[Ca^{2+}]}_{\text{final}} = \dfrac{\rm{K}_{\text{sp}}} {[\ce{F^-}]^2}= \dfrac{1.7 \times 10^{-10}}{0.0501^2} = 6.77 \times 10^{-8}$

$\text{moles }\ce{Ca^{2+}} \text{ final}= (6.77 \times 10^{-8})(1.946) = 1.32 \times 10^{-7}$

% $\ce{Ca^{2+}}$ recovered = $(1-\dfrac{1.32 \times 10^{-7}}{2\times 10^{-5}}) \times 100\% = 99.34 \%$ 

So we get a slightly different value, but it still seems like a magic number of some sort. 



Now if I were to write the problem I would have stated:


  You are given $1.00$ Litre of solution which is saturated with  $\ce{CaF_2}$ and has a $\ce{Ca^{2+}}$ concentration of $2.00 \times 10^{-5}$ M. How many mililitres of $0.100$ M $\ce{NaF}$ would you need to recover 99% of the $\ce{Ca^{2+}}$ as a precipitate of $\ce{CaF_2}$? The solubility product of $\ce{CaF_2}$ is $1.7\times10^{-10}$ .




Curious I played with this some more using the assumption of the $\ce{F^-}$ saturated starting solution. Obviously if you add vast amounts of $\ce{F^-}$ solution then any $\ce{CaF_2}$ which formed would dissolve agin. So there is some amount of $\ce{F^-}$ solution which precipitates the maximum amount of $\ce{CaF_2}$.

The max % $\ce{Ca^{2+}}$ recovery = 99.44% with 1913 ml of $\ce{F^-}$ solution. To +/- 0.01% any volume between about 1400 ml and 2500 ml would yield effectively the same result.