Is this problem about acid-base titration wrong?
Your calculations are correct. I also did a second sanity check: Adding $200~\mathrm{ml}\ \ce{NaOH}$ to a neutral solution does indeed give a $\mathrm{pH}\ 12$ solution, so they also wanted you to assume that the neutral point is the equivalence point.

Which answer to select? Well, considering that:

$$-\log(2 \cdot 10^{-2}) \approx 1.699 \approx 2$$

I would probably go with $2$. Along with telling the pupils that this is a badly worded question and in their exam there will be clearly solveable ones.