Answer
$14.8g\space C_3H_8O$
Work Step by Step
We know that
$\Delta T_f=K_fm$
This can be rearranged as:
$m=\frac{\Delta T_f}{K_f}$
We plug in the known values to obtain:
$m=\frac{1.50}{1.86}=0.806\frac{mol}{Kg}$
Now $0.200Kg\space H_2O\times \frac{0.806mol\space C_3H_8O_3}{1Kg\space H_2O}\times \frac{92.09g\space C_3H_8O_3}{1mol\space C_3H_8O_3}=14.8g\space C_3H_8O$