For vaporisation, $\Delta H = \Delta U + \Delta n_g RT$ where $\Delta n_g$ is the change in moles of gaseous species.

In vaporisation of $1$ mol water: $\mathrm{H_2O(l) \to H_2O(g)}$ Liquid is not counted in gaseous moles, so $\Delta n_g = 1 - 0 = 1$

Therefore, $\Delta U = \Delta H - RT$

Given: $\Delta H = 40.79\ \text{kJ mol}^{-1}$ $T = 100^\circ C = 373\ \text{K}$ $R = 8.3\ \text{J K}^{-1}\text{mol}^{-1}$

Calculate $RT$: $RT = 8.3 \times 373 = 3095.9\ \text{J mol}^{-1} = 3.096\ \text{kJ mol}^{-1}$

Now, $\Delta U = 40.79 - 3.096 = 37.694\ \text{kJ mol}^{-1}$

Since integer answer is required, $\Delta U \approx 38\ \text{kJ mol}^{-1}$