Given :

$\begin{array}{r}\left(\mathrm{I}\right){\mathrm{H}}_2\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)\to {\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\\ {\mathrm{\Delta H}}_{298\mathrm{K}}^{\circ }=-285.9\mathrm{kJ}{\mathrm{mol}}^{-1}\end{array}$

$\begin{array}{l}\left(\mathrm{II}\right) {\mathrm{H}}_2\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)\to {\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)\\ {\mathrm{\Delta H}}_2^{\circ }298\mathrm{K}=-241.8\mathrm{kJ}{\mathrm{mol}}^{-1}\end{array}$

The molar enthalpy of vapourisation of water will be :

$\begin{array}{l}{\mathrm{H}}_2\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)⟶{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\\ {\mathrm{\Delta H}}^{\circ }=-285.9 \mathrm{kJ} {\mathrm{mol}}^{-1} . . . .\left(1\right)\end{array}$

${\mathrm{H}}_2\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)⟶{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

${\mathrm{\Delta H}}^{\circ }=-241.8 \mathrm{kJ} {\mathrm{mol}}^{-1}$              . . . . .(2)

We have to calculate

${\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)⟶{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right);{\mathrm{\Delta H}}^{\circ }=?$

On substracting eqn. (2) from eqn. (1) we get

${\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)⟶{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

${\mathrm{\Delta H}}^{\circ }=-241.8-(-285.9)$

$=44.1 \mathrm{kJ} {\mathrm{mol}}^{-1}$