`PV=(w)/(m)RT` (for vapours of `H_(2)O`)
`P=3.6xx10^(3)Pa`, `V=2xx10^(-3)m^(3)`, `T=300 K`
`:. W_(H_(2)O)=(3.6xx10^(3)xx18xx2xx10^(-3))/(8.314xx300=0.052)`
`w_(H_(2)O)=52 g`
Since, realtive humidity `=60%`, therefore, amount of `H_(2)O=52xx0.6=31.2 g`