CaSO4 precipitate in mixture of Na2SO4 and CaCl2 solutions
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.