The vapor pressure of a solution at $100°\text{C}$  containing 3g of cane sugar in 33g of water (vapor pressure of water at $100°\text{C}=\text{760}\text{mm}$ of Hg) is

Weight of cane sugar $\text{w}=3\text{g}$  

Molar mass of cane sugar, $\text{M}\text{.W}=342\text{g/mol}$

Weight of water (solvent), $\text{W}=33\text{g}$

Vapor pressure of pure solvent (water),

$\text{P}°=760\text{mm}\text{of}\text{Hg}$

Vapor pressure of solution, $\text{P}=x$

From the Raoult’s law,

$\frac{\text{P}°-\text{P}}{\text{P}°}=\frac{\text{w}}{\text{m}}\times \frac{\text{M}}{\text{W}}$

$\frac{760-x}{760}=\frac{3}{342}\times \frac{18}{33}$

$761-x=\frac{760}{19\times 11}=3.63$

$x=760-3.63$

$\therefore \text{Vapour}\text{pressure}\text{of}\text{solution, }x=756.36\text{mm of Hg}$