When you don't have steam tables, an approximation of the water liquid-vapour equilibrium is P(atm) = [0,01*T(°C)]4 between 1atm and a few 10atm. Tables are obviously better.
The question says "rigid tanks" so I feel no need to compute a final density. Each chamber keeps its volume and contents, heat just flows from one to the other, and the heat capacity at constant volume (or better, the tabulated internal heat) tells the final temperature. The only subtlety I see is that the masses differ, but you had already caught it.
As a quick check, I take a constant CV (hence the same for both chambers) and get a final temperature of 268°C so 251°C might be possible with detailed data. On the other hand, 164°C isn't reasonable for being too close to one initial temperature and not even between both.