Answer
2.5 kg.
Work Step by Step
The saturated vapor pressure of water at $20^{\circ}$ is 2330 Pa, from Table 13–3.
Use the ideal gas law and find the maximum amount of water vapor that can be in the air.
$$PV=nRT$$
$$n=\frac{PV}{RT}=\frac{(2330Pa)(420m^3)}{(8.314J/mol \cdot K)(293K)}=401.7 moles$$
The relative humidity is 65 percent, so the air could still hold 35 percent of the total possible water. Find the mass.
$$m=0.35(401.7moles)\frac{0.018kg}{mole}=2.5kg$$