V = Volume available to the gas
n = Number of moles of gas
R = ideal gas constant
T = temperature in kelvin
Now, the total pressure in the problem is from a mixture of gasses, and we're only interested in the water that escaped from a liquid solution into the vapor phase. To solve this based on given information, we're going to have to make some assumptions. First, let's assume that no other molecules collected, except for water, were due to them evaporating (i.e. they didn't escape into the vapor phase). Second, let's assume that the atmospheric pressure where the gas sample was collected is exactly 1 atm. Thus, the partial pressure due to atmospheric gasses would be 1 atm, and the partial pressure due to escaped water vapor would be 0.198 atm.
Next, let's write the significant variables in the same units as they appear in the ideal gas constant we're using:
P (escaped H2O) = 0.198 atm
V = 0.072 L
T = 287.15 K
Now, we solve the ideal gas equation for n, number of moles, then plug in our values to get the number of moles:
n = P*V/(R*T) = 6.050 * 10^(-4) mol
Now, we just convert mols of water to grams using the molar mass of water, which gives us 0.0109 grams of water.