We know,

$$\Delta {T_f}\, = \,im{K_f}$$

Where

$$\Delta {T_f}\, = \,$$ Depression in freezing point

i = no of ions or molecule

m = molality of solute

K_f = freezing point depression constant.

In this question m = 1 and K_f is constant.

So, $$\Delta {T_f}\,\, \propto \,\,i$$

Which solutions have more i, those solutions $$\Delta {T_f}\,$$ will be more.

Now,

Freezing point of a aqueous solution $${f_p} = \, - \,\Delta {T_f}$$

So, those solutions have more $$\Delta {T_f}$$ will have less freezing point

and $$\Delta {T_f}$$ will be more when value of i is more.

Now, for,

[ Co (H₂O)₆] Cl₃ $$\rightleftharpoons$$ [ Co(H₂O)₆]³⁺ + 3Cl^$$-$$

Value of i = 4 (no. of ions), 3 ions of Cl^$$-$$

and one ion of [ Co (H₂O)₆]³⁺

[ Co (H₂O)₅ Cl ] Cl₂ . H₂O $$\rightleftharpoons$$ [ Co (H₂O)₅ Cl ]²⁺ + 2Cl^$$-$$

Value of i = 3

[ Co (H₂O)₄ Cl₂ ] Cl . 2H₂O $$\rightleftharpoons$$ [ Co (H₂O)₄ Cl₂ ] ⁺ + Cl^$$-$$

Value of i = 2

and [ Co(H₂O)₃ Cl ] . 3H₂O do not show ionization.

So, i = 1.

As, [ Co (H₂O)₃ Cl ]. 3H₂O have lowest value of i,

So, it's freezing point will be maximum.