For gaseous water PV = nRT
Thus Volume occupied by 1 mole gaseous water can be derived as
T = 273 + 100 = 373 K
$V = \large\frac{nRT}{P} = \large\frac{1\times0.0821\times373}{1}$
$\;\;\;\;\;\;\;\;\;\;\; = 30.62 litre$
Also Volume of 1 mole of liquid water = $\large\frac{mass}{density}$
$ = \large\frac{18}{0.958}$
=18.79 mL
$=18.79\times10^{-3}$ litre
Thus Volume Percentage occupied by water molecules in gaseous state
$=\large\frac{18.79\times10^{-3}}{30.62}\times100$
= 0.0614
$\therefore \%$ of free volume = 100 - 0.0614
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=99.9386$
Hence answer is (b)