At room temperature a dilute solution of urea is prepared by dissolving 0.60g of urea in 360g of water. If the vapour pressure of pure water at this temperature is 35mm Hg, Lowering of vapour pressure will be ($molar mass of urea = 60g mo{l}^{–1}$)

Given data:
It is given to identify  the lowering of vapour pressure. 
Explanation:
$$\begin{gathered} Vapour pressure reduction = {\mathrm{P}}^0-\mathrm{p}={\mathrm{p}}^0·{\mathrm{x}}_{\text{solute }} \\ = \mathrm{\Delta P}=35\times \frac{0.6/60}{\frac{0.6}{60}+\frac{360}{18}} \\ = 35\times \frac{.01}{.01+20}=35\times \frac{0.01}{20.01} = =0.017\mathrm{mmHg} \end{gathered}$$

Hence, the correct option is option (D) 0.017 mmHg.