molar solubility of CaF2 at pH3 given molar solubility at pH7 and pKa of HF
The $K_{sp}$ of $\ce{CaF2}$ which you found out as $3.7 \times 10^{-11}$ is indeed correct, and after that essentially you have to solve two equations. 

Now, suppose at $pH =3$, (i.e. $\ce{[H+] = 10^{-3}}$),  $\ce{CaF2}$ dissociates as $\ce{Ca^2+}$ and $\ce{F-}$ and their concentration in the solution are as follows, $$\ce{CaF2 &lt;=&gt; Ca^2+ + 2F-}$$ $\ce{[Ca^2+ ] = S }$ mol/L. and $\ce{[F-] = 2S -}$ $x$ mol/L. ,as, this $x$ mol/L. of $\ce{F-}$ is in equilibrium with $\ce{HF}$, as, $$\ce{H+ + F- &lt;=&gt; HF}$$ This equilibrium satisfies the equation $$\ce{\frac{[H^+][F^-]}{[HF]}= 10^{-3.17 }   =&gt; \frac{10^{-3}(2S - x)}{x} = 10^{-3.17}  ...(i)}$$.

And another equation which you have to solve is just regarding $K_{sp}$, i.e $$S(2S-x)^2 = 3.7 \times 10^{-11} .....(ii)$$ From first equation, you will get $\frac{2S}{x} = 1.676$ ; if you substitute for $x$ in the second equation by this, you can now solve for $S$ and the answer will be $S= 3.845 \times 10^{-4}$.  

Thus , the factor will be $\frac{3.845 \times 10^{-4}}{2.1 \times 10^{-4}} = 1.83$