Answer
226 kJ
Work Step by Step
$q=n \Delta H_{vap}$ where $n$ is the number of moles and $\Delta H_{vap}$ is the heat of vaporisation.
That is,
$q=100.0\,mL\times\frac{1.00\,g}{mL}\times\frac{1\,mol\,H_{2}O}{18.02\,g\,H_{2}O}\times\frac{40.7\,kJ}{1\,mol\,H_{2}O}$
$=226\,kJ$
