Fluoride concentration in LiF and CdF2 solutions Trying to learn how to use $\ce{K_{\mathrm{sp}}}$ values and not understanding. See question below:

Two saturated aqueous solutions are prepared at $\pu{25^\circ C}$
Solution1 is made by dissolving cadmium fluoride ($\ce{CdF2}$, $\ce{K_{sp}} = \pu{6.44E−3}$) in $\pu{100.0 mL}$ of water until excess solid is present.
Solution 2 is prepared by dissolving lithium fluoride ($\ce{LiF}$, $\ce{K_{sp}} = \pu{1.84E−3}$) in $\pu{200.0 mL}$ of water until excess solid is present.
Which solution has a higher fluoride ion concentration?&quot;

I used the $\ce{K_{sp}}$ given for $\ce{CdF2}$ to get $s = 0.11720$, so $\ce{[F]} = 2s = \pu{0.2344 M}$, is that right? I guess I'm not sure what to do after that to get find the concentration in $\pu{100 mL}$.  Wouldn't it be $\pu{2.344 M}$?  I am told this answer is incorrect.
The problem as stated is not asking you which solution has the higher amount of fluoride but which one has the higher concentration of fluoride, i.e. the higher molarity (as this is usually the concentration unit more often used by chemists).
You need to calculate the fluoride concentration for the $\ce{LiF}$ solution in the same way you did for the $\ce{CdF2}$ solution. My answer comes to 0.04290 M.
Thus, the $\ce{CdF2}$ solution has a fluoride concentration of 0.2344 M while the $\ce{LiF}$ solution has a fluoride concentration of 0.04290 M. Therefore, the answer to the problem is: the $\ce{CdF2}$ has a higher fluoride ion concentration.