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.
