The first thing that you need to do here is to calculate the molality of the solution.
To do that, you can use the equation
#color(blue)(ul(color(black)(DeltaT = i * K_b * b)))#
#DeltaT#is the boiling point elevation, measured as the difference between the boiling point of the solution and the boiling point of the pure solvent
#i#is the van't Hoff factor, equal to #1#for non-ionizing solutes
#K_b#is the ebullioscopic constant of the solvent, equal to #"2.65 K kg mol"^(-1)#for benzene #->#see here
#b#is the molality of the solution
Now, you know that when you dissolve
#DeltaT = "0.531 K"#
Rearrange the equation to solve for
#DeltaT = i * K_b * b implies b= (DeltaT)/(i * K_b)#
Plug in your values to find
#b = (0.531 color(red)(cancel(color(black)("K"))))/(1 * 2.65 color(red)(cancel(color(black)("K"))) "kg mol"^(-1)) = "0.2004 mol kg"^(-1)#
This tells you that your solution contains
You can thus say that your sample will contain
#100 color(red)(cancel(color(black)("g benzene"))) * "0.2004 moles solute"/(10^3color(red)(cancel(color(black)("g benzene")))) = "0.02004 moles solute"#
Finally, to find the molar mass of the solute, simply divide its mass by the number of moles it contains.
#M_ ("M solute") = "2.67 g"/"0.02004 moles" = color(darkgreen)(ul(color(black)("120 g mol"^(-1))))#
I'll leave the answer rounded to two sig figs, but keep in mind that you have one significant figure for the mass of benzene.