Given : $L_f=80 cal g^{-1},L_{\nu}=540cal g^{-1}$
We know that,
$\Delta T_b=K_b\times $ molality
$\Delta T_f=K_f\times $ molality
Also,$K=\large\frac{RT^2}{1000\times L}$
$\therefore \large\frac{\Delta T_b}{\Delta T_f}=\frac{K_b}{K_f}$
$\Rightarrow \large\frac{RT_b^2}{1000\times L_{\nu}}\times \frac{1000\times L_f}{RT_f^2}$
$\large\frac{\Delta T_b}{\Delta T_f}=\frac{T_b^2\times L_f}{T_f^2\times L_{\nu}}$
$\large\frac{0.1}{\Delta T_f}=\frac{373\times 373\times 80}{273\times 273\times 540}$
$\therefore \Delta T_f=0.361$
Freezing point $(T_f)=0-\Delta T_f=0-0.361$
$\Rightarrow -0.361^{\large\circ}C$