Elevation in boiling point of water
While considering colligative properties we always assume solute to be non-volatile. So, it doesn't contribute to the vapor. The amount of vapor will only be due to solvent particles.
So, in the problem we can use the following relation:
$$ \Delta T = K_\mathrm b\cdot m $$ 
Initially $ \Delta T  = 0.53\ \mathrm{^\circ C}$ so plugin the value of $K_\mathrm b$ and you end up with the value of initial molality of solvent as $1$. From this you can find the initial moles of solute which remain constant throughout. 

Next you put $ \Delta T  = 2.12\ \mathrm{^\circ C}$ and plug in the known values of $K_\mathrm b$ and 'moles of solute' in the above formula to get the final value of solvent amount that remains in the solution.
Subtracting the final amount of solvent left from the initial amount will let you know the amount of water vaporized which is $750\ \mathrm g$.