Increasing boiling point.....
You got the right order, but for the wrong reasons. CH3Br is more polar than CHCl3, but CHCl3 has the greater boiling point because of London dispersion forces.
The liquid with highest boiling point will have the strongest intermolecular attractions. Intermolecular attractions include hydrogen bonding and the three van der Waals forces. The three van der Waals forces are Keesom forces (dipole-dipole attraction), Debye forces (induced attraction), and London dispersion forces (all molecules exhibit these). Interestingly, London dispersion forces are often greater than either Keesom forces or Debye forces and are second only to hydrogen bonding. The strength of London dispersion forces is proportional to the total number of electrons and to the surface area over which they are spread. Many textbooks, and some teachers' dusty old notes relegate London dispersion forces to the bottom of the pile when it comes to intermolecular attraction. That is a mistake.
-------------- increasing boiling point ------->
lowest ................... ........................highest
Cl2 ....... CH3Br ......... CHCl3 ........ H2O
Indeed, H2O is the highest since it exhibits hydrogen bonding and all three van der Waals forces. Cl2 is the lowest since it is nonpolar, and only exhibits London dispersion forces.
That leaves chloroform, CHCl3, and methyl bromide. We predict that CH3Br is probably more polar, but the greater London dispersion forces of CHCl3 will give it a greater boiling point.
Lets see how our prediction stacks up. CHCl3 is chloroform and it is a liquid at room temperature and boils at 61C. Methyl bromide boils about 4C. The net dipole moments CH3Br and CHCl3 are 1.82D and 1.15D, respectively. You might expect CH3Br to have the greater boiling point because it is more polar, but it doesn't. The greater London dispersion forces of CHCl3 will give it a greater boiling point than CH3Br. This should provide the evidence you need to realize that London dispersion forces are often stronger than Keesom forces (dipole-dipole attraction).
Next, consider CO2, CH3CH3, CH3CH2OH, and (CH3)2O. We know CO2 is a nonpolar gas with a low boiling point if the pressure is much greater than atmospheric pressure. Remember, CO2 undergoes sublimation at 1.00 atm, it does not form a liquid except at pressures above about 5 atm. Therefore, we can kick CO2 out of the list because it doesn't have a boiling point (at 1.00 atm). We know that ethanol, CH3CH2OH is a liquid at room temperature and exhibits hydrogen bonding, so it will have the highest boiling point. C2H6 (18 electrons) is nonpolar and only exhibits London dispersion forces, and will have the lowest boiling point. That leaves dimethyl ether as the one in the middle since it is "bent" and therefore polar and so it exhibits all three van der Waals forces and boils at a higher temperature than ethane.
Let's see how we did: C2H6.... -88.5C ......... CH3OCH3 .... -24C .......... ethanol .... 78C
The dipole moment of CH3OCH3 is 1.30D. So indeed, our prediction that it will have a net dipole moment because it is a bent molecule is correct.