To determine which solution has the lowest freezing point, you should consider the van 't Hoff factor $i$, which represents the number of particles the solute dissociates into when dissolved. The freezing point depression is given by:

$$ \Delta T_f = i \cdot K_f \cdot m $$

where:

$\Delta T_f$ is the change in freezing point,

$K_f$ is the freezing point depression constant,

$m$ is the molality of the solution,

$i$ is the van 't Hoff factor, which depends on the dissociation of the solute.

For each solute, the van 't Hoff factor $i$ can be determined as follows:

Option A: $\text{Al}_2(\text{SO}_4)_3$

Dissociates into 2 Al$^{3+}$ ions and 3 SO$_4^{2-}$ ions.

$i = 2 + 3 = 5$

Option B: $\text{C}_6\text{H}_{12}\text{O}_6$ (glucose)

Does not dissociate in solution (non-electrolyte).

$i = 1$

Option C: $\text{KI}$

Dissociates into K$^+$ and I$^-$ ions.

$i = 1 + 1 = 2$

Option D: $\text{K}_2\text{SO}_4$

Dissociates into 2 K$^+$ ions and 1 SO$_4^{2-}$ ion.

$i = 2 + 1 = 3$

Comparing the van 't Hoff factors, the solution of $\text{Al}_2(\text{SO}_4)_3$ will have the highest factor ($i = 5$), leading to the greatest freezing point depression. Thus, the $\text{Al}_2(\text{SO}_4)_3$ solution will have the lowest freezing point since the extent of freezing point depression is directly proportional to $i$.

Therefore, the solution of $\text{Al}_2(\text{SO}_4)_3$ has the lowest freezing point.