The molar mass is 49.8 g/mol..

The formula for freezing point depression is

#color(blue)(bar(ul(|color(white)(a/a)ΔT_"f" = K_"f"bcolor(white)(a/a)|)))" "#

where

#ΔT_"f"# = the decrease in the freezing point

#K_"f"# = the molal freezing point depression constant of the solvent

#b# = the molality of the solution

We can rearrange the formula to get

#b = (ΔT_"f")/K_"f"#

In your problem,

#ΔT_"f" = "3.12 °C"#

#K_"f" = "1.86 °C·kg·mol"^"-1"#

∴ #b = (3.12 color(red)(cancel(color(black)("°C"))))/(1.86 color(red)(cancel(color(black)("°C")))·"kg·mol"^"-1") = "1.677 mol·kg"^"-1"#

Now, #b = "1.677 mol"/(1 color(red)(cancel(color(black)("kg")))) = "0.8763 g"/("0.010 4832" color(red)(cancel(color(black)("kg"))))#

∴ #"1 mol" = "0.8763 g"/"0.010 4832" × 1/1.677 = "49.8 g"#

The molar mass is 49.8 g/mol.