The vapor above the mixture will consist of both ethanol and water molecules because both components of the mixture are volatile in this case. Therefore, Vapor pressure of the mixture=vapor pressure of ethanol above the mixture + vapor pressure of water above the mixture. Now, simply use the vapor pressure lowering equation (X*Po=P) for ethanol, then water, and add them together. Po, the vapor pressure of pure water or pure ethanol, is given in the problem. X, the mole fraction, is solvable since you know the mass values of ethanol and water.