Molality =0.1539m
$\Delta T_f=273.15-271K$
$\qquad=2.15K$
$\Delta T_f=K_f.m$
$m=\large\frac{\Delta T_f}{K_f}$
$0.1539=\large\frac{2.15}{K_f}$
$K_f=\large\frac{2.15}{0.1539}$
$W_2=5g$
$M_2=180gmol^{-1}$
$\Delta T_f=K_f\times m$
$\Rightarrow \large\frac{2.15}{0.1539}\times \frac{5}{180}\times \frac{1000}{95}$
$\Rightarrow 4.08K$
Freezing point of solution =273.15-4.08K
$\Rightarrow 269.27K$