The formula for boiling point elevation #ΔT_"b"# by a nonvolatile solute is

#color(blue)(bar(ul(|color(white)(a/a)ΔT_"b" = K_"b"mcolor(white)(a/a)|)))" "#

where

#K_"b"# = the molal boiling point elevation constant of the solvent

#m# = the molality of the solution

We can rearrange the formula to get

#m = (ΔT_"b")/K_"b"#

In your problem,

#ΔT_"b" = "0.579 °C"#

#K_"b" = "5.02 °C·kg·mol"^"-1"#

∴ #m = (0.579 color(red)(cancel(color(black)("°C"))))/(5.02 color(red)(cancel(color(black)("°C")))·"kg·mol"^"-1") = "0.1153 mol·kg"^"-1"#

Now, #m = "0.1153 mol"/(1 color(red)(cancel(color(black)("kg")))) = "0.25 g"/("0.0400" color(red)(cancel(color(black)("kg"))))#

Divide both sides of the equation by #0.1153#.

∴ #"1 mol" = "0.25 g"/0.0400 × 1/0.1153 = "54 g"#

The molar mass is 54 g/mol.