Given, ${\text{H}}_2\mathrm{O}\left(\mathrm{l}\right)\rightleftharpoons {\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)\hspace{0.17em}\mathrm{at}\hspace{0.17em}373\mathrm{\hspace{0.17em}}\mathrm{K},\mathrm{\hspace{0.17em}}{\mathrm{\Delta H}}^{\mathrm{o}}=8.31 \mathrm{kcal} {\mathrm{mol}}^{-1}$

Thus, boiling point of 0.1 molal sucrose solution is 

${\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$ changes to steam ${\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$ at 373 K. Thus, this represents latent heat of vaporisation. ${\mathrm{K}}_{\mathrm{b}}$ (molal elevation constant) is related to ${\mathrm{\Delta H}}^{\mathrm{o}}$ and boiling point by equation, 

${\mathrm{K}}_{\mathrm{b}}=\frac{{\mathrm{RT}}_0^2}{1000{\mathrm{\Delta H}}^{\mathrm{o}}}$

Here, ${\mathrm{\Delta H}}^{\mathrm{o}}$ is energy unit per gram of solvent.

${\mathrm{\Delta H}}^{\mathrm{o}}=8.31 \mathrm{kcal} {\mathrm{mol}}^{-1}$

          $=\frac{8.31}{18}\mathrm{kcal} {\mathrm{g}}^{-1}$

${\mathrm{K}}_{\mathrm{b}}=\frac{0.002\times {\left(373\right)}^2}{1000\times \left(\frac{8.31}{18}\right)}$

     $=\frac{278.25}{461.66}=0{.60}^{\mathrm{o}} {\mathrm{mol}}^{-1}\mathrm{kg}$

${\mathrm{\Delta T}}_{\mathrm{b}}$ (sucrose solution) = molality $\times {\mathrm{K}}_{\mathrm{b}}=0.1\times 0.60=0{.06}^{\mathrm{o}}$

$\therefore$ Boiling point of solution $=373+0{.06}^{\mathrm{o}}=373.06\mathrm{K}$