I don't like this question.
First of all - at room temperature butane is already a gas, it boils around 0°C.
Then, rate of evaporation depends on many factors, one of them being enthalpy of vaporization. Assuming all other factors to be identical, it is the compound with lower enthalpy of vaporization that will evaporate faster. According to NIST database methanol has enthalpy of vaporization around 38 kJ/mol, butane has enthalpy of vaporization around 22 kJ/mol - so you are right it is D that seems to be the correct answer.
I wonder if (whoever asked the question) didn't think in terms of "lower molar mass, faster evaporation" (which is not a bad starting point), but forgot to account for the hydrogen bonding, which is a show stopper in methanol.