$\begin{array}{l}\mathrm{KBr}=\mathrm{K}+{\mathrm{Br}}^{-}=2 \mathrm{ions}\\ {\mathrm{BaCl}}_2={\mathrm{B}}^{2+}+2{\mathrm{Cl}}^{-}=3 \mathrm{ions}\end{array}$

${\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3\to 2{\mathrm{Al}}^{+3}+3{\mathrm{SO}}_4^{-2}\Rightarrow 5 \mathrm{ions}$

$\therefore$ urea is not ionised hence it is shows highest freezing point. 

$∆{\mathrm{T}}_{\mathrm{f}} \mathrm{\alpha } \frac{1}{\mathrm{Freezing} \mathrm{point} \mathrm{of} \mathrm{solution}}$