Gibbs free energy for the evaporation of water
The average heat capacity $C_p$ of liquid water and water vapor over the range between $\pu{25^\circ C}$and $\pu{100^\circ C}$ are, respectively $4.18\ \pu{kJmol^{-1}K^{-1}}$ and $1.72\ \pu{kJmol^{-1}K^{-1}}$.  So the enthalpies of liquid water and water vapor at $\pu{100^\circ C}$ are:
$H_l(\pu{100^\circ C})=-285.8+(0.018)(4.18)(75)=-280.2\ \pu{kJmol^{-1}}$
$H_g(\pu{100^\circ C})=-241.8+(0.018)(1.72)(75)=-239.5\ \pu{kJmol^{-1}}$
So, at $\pu{100^\circ C}$, $\Delta H=40.7\ \pu{kJmol^{-1}}$
The entropies of liquid water and water vapor at $\pu{100^\circ C}$ are:
$S_l(\pu{100^\circ C})=0.06996+(0.018)(4.18)\ln{(373.15/298.15)}=0.08684\  \pu{kJmol^{-1}K^{-1}}$
$S_g(\pu{100^\circ C})=0.1887+(0.018)(1.72)\ln{(373.15/298.15)}=0.1956\  \pu{kJmol^{-1}K^{-1}}$
So, at $\pu{100^\circ C}$, $\Delta S = 0.1088\ \pu{kJmol^{-1}}$ and $T\Delta S=40.6\ \pu{kJmol^{-1}}$ So the change in Gibbs free energy between saturated liquid and saturated vapor at $\pu{100^\circ C}$ and 1 atm is $$\Delta G=40.7-40.6=0.1\ \pu{kJmol^{-1}}$$ which, to within roundoff error is zero (as expected).
The Spoiler answer is obviously incorrect.