acid base - Q. 36 2018 molar solubility of CaF2 at pH3 given molar solubility at pH7 and pKa of HF

This is Q36 on the 2018 Chemistry Olympiad:

Calcium fluoride, $\ce{CaF2}$, has a molar solubility of $\pu{2.1e–4 mol L–1}$ at pH = 7.00. By what factor does its molar solubility increase in a solution with pH = 3.00? The pKa of $\ce{HF}$ is 3.17 (source).

Here are my steps:

1. Calculate the solubility product of calcium fluoride from the data at pH=7.



2. Combine the dissolution and the acid/base reaction to $$\ce{CaF2(s) + 2 H+ <=> Ca^2+(aq) + 2HF(aq)}$$


3. Calculate the equilibrium constant of the combined reaction from the known constants of the separate reactions.


4. Calculate the equilibrium concentration of $\ce{Ca^{2+}}$ at pH = 3.00


5. Calculate the molar solubility and compared it to the one given for ph = 7.

The arithmetic is shown on the picture, resulting in a $\ce{Ca^{2+}}$ concentration of $\pu{4.325e-4 mol L-1}$.

I don't understand what I did wrong as the answer is B, 1.83.

Basically I combined the equilibrium of $\ce{CaF2}$ dissociating and the equilibrium of the formation of $\ce{HF}$. I then using the equilibrium product found out the concentration of $\ce{Ca^{2+}}$. It would be great if you could show me where I went wrong.