The mass of glucose that should be dissolved in $50 \mathrm{g}$ or water in order to produce the same lowering of vapour pressure as is produced by dissolving $1 \mathrm{g}$ of urea in the same quantity of water is

$\frac{P^{\circ }-P_s}{P_s}=\frac{w_1{M}_2}{w_2{M}_1}$

To produce same lowering of vapour pressure $\left(\frac{P^{\circ }-P_s}{P_s}\right)$ term will be same for both the cases 

(solvent is same) 

$\begin{array}{l}\therefore \frac{w_g\times 18}{50\times 180}=\frac{w_u\times 18}{50\times 60}\\ \left(w_g\right.=\text{ weight of glucose },w_u=\text{ weight of urea) }\\ or,\frac{w_g\times 18}{50\times 180}=\frac{1\times 18}{50\times 60}\text{ or }{w}_g=3.\end{array}$