Relative lowering of vapour pressure$=$Mole fraction of solute

$\frac{\mathrm{Lowering} \mathrm{of} \mathrm{vapour} \mathrm{pressure}}{\mathrm{Vapoure} \mathrm{pressure} \mathrm{of} \mathrm{solution}}=\frac{\mathrm{moles} \mathrm{of} \mathrm{solute}}{\mathrm{moles} \mathrm{of} \mathrm{solvent}}$

$\frac{25}{75}=\frac{\text{ moles of urea }}{\left(\frac{1000}{18}\right)}$

$\frac{1}{3}\times \frac{1000}{18}=$moles of urea

Mass of urea$=\frac{1}{3}\times \frac{1000}{18}\times 60$

$=1111.11 \mathrm{gm}$