6 g of urea is dissolved in 90g of boiling water. The vapour pressure of the solution is 

 We know vapour pressure can be mathematically represented as:
 $\mathrm{P}=\mathrm{P}°\mathrm{x}$

where $\mathrm{P}°$ represents the boiling point of pure solvent.
$$\begin{gathered} P=\mathrm{P}°\frac{{\mathrm{n}}_1}{{\mathrm{n}}_1+{\mathrm{n}}_2} \\ =760\left(\frac{0.1}{0.1+5}\right) 90{\mathrm{gH}}_2\mathrm{O}=5 \mathrm{mol}={\mathrm{n}}_2; \text{ g urea }=0.1 \mathrm{mol}={\mathrm{n}}_1 \\ =760\left(\frac{0.1}{0.1+5}\right) \\ =760\times 0.0196=14.89\text{ Torr }=\text{ lowered pressure } \end{gathered}$$
Now, 1 torr $=1 \mathrm{mm}$ of $\mathrm{Hg}$
Vapour pressure of solution $=760-14.89\approx 744.8 \mathrm{mm}$ of $\mathrm{Hg}$
So the required solution is Option A) 744.8 mm of Hg.