CaSO4 precipitate in mixture of Na2SO4 and CaCl2 solutions I have been stuck on a chemistry problem for a long time now and if anyone here can help me I would be eternally grateful.

40. Du tillsätter $\pu{100 ml}$ $\pu{0,01 M}$ $\ce{Na2SO4}$-lösning i $\pu{100 ml}$ $\pu{0,02 M}$ $\ce{CaCl2}$-lösning. Det bildas en $\ce{CaSO4}$-fällning. Hur mycket av $\pu{0,01 M}$ $\ce{Na2SO4}$-lösning måste ännu tillsättas för att den bildade $\ce{CaSO4}$-fällningen just och just ska upplösas helt? $K_\mathrm{s}(\ce{CaSO4}) = \pu{2,5e-5 M}$. $(t = \pu{25 °C})$ (2p)
a. $\pu{0,43 l}$
b. $\pu{0,48 l}$
c. $\pu{0,53 l}$
d. $\pu{0,58 l}$

Translation: We add 100 ml of 0.01 M $\ce{Na2SO4}$-solution to 100 ml of 0.02 M $\ce{CaCl2}$-solution. A precipitate of $\ce{CaSO4}$ is formed. What is the (minimal) volume of 0.01 M $\ce{Na2SO4}$-solution that needs to be added to the mix for the $\ce{CaSO4}$ precipitate to be just dissolved completely?
The problem then lists the solubility constant, $K_\mathrm{s}$, for $\ce{CaSO4}$ as $2.5\cdot 10^{-5}$.
The only way I can think of solving this myself is by looking at the added $\ce{Na2SO4}$-solution as just water and using the $K_\mathrm{s}$ value for $\ce{CaSO4}$ to see what concentration of $\ce{CaSO4}$ is possible in just water. Then calculate the amount of moles of precipitate that was actually formed to see how much water I would need to add to the mixture:
$$K_\mathrm{s}(\ce{CaSO4}) = \pu{2.5e-5 M^2} = [\ce{Ca^2+}][\ce{SO4^2-}]$$
$$\therefore [\ce{Ca^2+}] = [\ce{SO4^2-}] = \sqrt{\pu{2.5e-5 M^2}} = \pu{5e-3 M}$$
$$C = \frac{n}{V} \to n = CV$$
Amount of precipitate formed:
$$n(\ce{CaSO4}) = \pu{0.02 M} \cdot \pu{0.1 dm^3} = \pu{0.002 mol}$$
Required total volume of water to dissolve $\pu{0.002 mol}$ of $\ce{CaSO4}$:
$$V = \frac{n}{C} = \frac{\pu{0.002 mol}}{\pu{5e-3 M}} = \pu{0.4 dm^3}$$
This is wrong, but I can't figure out the right way to go about this. The answer surely has to do with how well $\ce{CaSO4}$ is dissolved in a $\ce{Na2SO4}$-solution as opposed to just water.
As you've already noted, there are 3 substances of interest here: $\ce{Ca^{2+}}, \ce{SO4^{2-}}, \ce{H2O}$.

Let $V$ be the additional volume of sodium sulfate solution. Then:

The total volume is $V + 0.2\ \mathrm{L}$.

The total amount of sulfate is $(V + 0.1\ \mathrm{L}) (0.01\ \mathrm{M})$.

The total amount of calcium is $(0.1\ \mathrm{L}) (0.02\ \mathrm{M})$.

We have then the following relationship:

$$\ce{[Ca^{2+}][SO4^{2-}]}=\frac{(0.1\ \mathrm{L}) (0.02\ \mathrm{M})}{V + 0.2\ \mathrm{L}}\frac{(V + 0.1\ \mathrm{L}) (0.01\ \mathrm{M})}{V + 0.2\ \mathrm{L}}\leq K_{\mathrm{sp}} = 2.5\times 10^{-5}$$

Now, it's just algebra to solve for the threshold value of $V$, keeping in mind that there are two solutions for $V$ because this is a quadratic equation. In this case, the correct solution is the positive root:

$$V \approx 0.48\ \mathrm{L}$$

The mistake you made is that the calcium and sulfate ion concentrations are definitely not equal, since the additional volume of solution contains only sulfate and not calcium.