Why is CH4 a gas at room temperature but CH3Cl a liquid?
Yes, I think that is the right answer expected for this question. 



Three factors to consider.

(1) Most importantly there four intermolecular forces which effect BP

Ionic bonds  > Ion-dipole interactions > H bonding >  dipole-dipole interactions > Van der Waals dispersion forces.

(2) Increasing the number of carbon atoms (e.g. n-alkanes) increases the BP

(3) Branching on carbon chain lowers the BP.



The answer being given though is a bit of hand-waving that oversimplifies the situation. Dipole moment alone can't explain the overall trends. Van der Waals dispersion forces must be considered also. 

For instance with increasing substitution $\ce{CH2Cl2}$, and $\ce{CHCl3}$ both have lower dipole moments than $\ce{CH3Cl}$ but higher BP's. $\ce{CCl4}$ has no dipole moment like methane, but has the highest BP of all. 

                                          Bond Lengths
           Dipole     B.P.       Mol.
           Moment     (°C)       Wt.       C-H     C-X
 CH4         0        −161.49    16.04     108.7    ---
 CH3Cl       1.9       −23.8     50.49     111      178.3   
 CH2Cl2      1.6        39.6     84.93     106.8    177.2
 CHCl3       1.15       61.15    119.37    107.3    176.7  
 CCl4        0          76.72    153.81    -----    176.6


Also look at the halomethanes as a series. $\ce{CH3F}$ has a smaller dipole moment than $\ce{CH3Cl}$ because the bond length is shorter. 

           Dipole   B.P      Mol.      Electro-  C-X     C-H
           Moment   (°C)     Wt.       Negat.    Length  Length
 CH3F       1.85    −78.4    34.03     4.0       138.5   109.5
 CH3Cl      1.87    −23.8    50.49     3.0       178.4   
 CH3Br      1.81      4.0    94.94     2.8       192.9
 CH3I       1.62      42     141.94    2.5       213.9 


The trend for molecular weight looks good, but the deuterated form of methane, $\ce{CD4}$, has the same (or at least nearly the same) boiling point as $\ce{CH4}$.