As per Hardy-Schulze formulation, the flocculation values of the following for ferric hydroxide sol are in the order:

We have to apply the Hardy-Schulze rule, which states first that “the coagulating (flocculating) power of an electrolyte for a given lyophobic sol is proportional to the valency of the ion carrying the charge opposite to that on the sol particles.”

For a ferric hydroxide sol the dispersed particles are $$\text{Fe(OH)}_3$$. In the preparation of this sol we write

$$\text{Fe}^{3+}+3\text{OH}^- \; \longrightarrow \; \text{Fe(OH)}_3\downarrow$$

The surface of each tiny precipitated particle retains a slight excess of $$\text{Fe}^{3+}$$ ions, so every sol particle is positively charged. Hence the counter-ions that can neutralise this charge—and therefore produce coagulation—must be anions. So, when we compare different electrolytes, we look only at the valency of their anions.

Let us list every electrolyte in the question together with the charge on its anion:

$$\begin{aligned} \text{K}_3[\text{Fe(CN)}_6] &\;\; \longrightarrow\; 3\text{K}^+ + [\text{Fe(CN)}_6]^{3-} &\quad (\text{anion charge } = -3)\\[6pt] \text{K}_2\text{CrO}_4 &\;\; \longrightarrow\; 2\text{K}^+ + \text{CrO}_4^{2-} &\quad (\text{anion charge } = -2)\\[6pt] \text{KBr} &\;\; \longrightarrow\; \text{K}^+ + \text{Br}^- &\quad (\text{anion charge } = -1)\\[6pt] \text{KNO}_3 &\;\; \longrightarrow\; \text{K}^+ + \text{NO}_3^- &\quad (\text{anion charge } = -1)\\[6pt] \text{AlCl}_3 &\;\; \longrightarrow\; \text{Al}^{3+} + 3\text{Cl}^- &\quad (\text{anion charge } = -1) \end{aligned}$$

According to Hardy-Schulze, the flocculation value $$F$$ (that is, the minimum millimoles of electrolyte required to coagulate one litre of sol) is inversely proportional to the coagulating power, and the coagulating power increases with anion valency. Symbolically,

Higher anion charge $$\; \Longrightarrow \;$$ higher coagulating power $$\; \Longrightarrow \; \text{smaller } F.$$

So we set up the chain

$$[ \text{Fe(CN)} _6]^{3-}\; (3-) \; \Longrightarrow \;$$ smallest $$F$$

$$\text{CrO} _4^{2-}\; (2-) \; \Longrightarrow \;$$ next smaller $$F$$

All monovalent anions $$\text{Cl}^- , \text{Br}^- , \text{NO}_3^-$$ have equal valency 1, so their coagulating power is the least and their flocculation value is the highest. Hence, within experimental error, we may write

$$F\bigl([\text{Fe(CN)}_6]^{3-}\bigr) \lt F\bigl(\text{CrO}_4^{2-}\bigr) \lt F\bigl(\text{Cl}^-\bigr)=F\bigl(\text{Br}^-\bigr)=F\bigl(\text{NO}_3^-\bigr).$$

Translating each anion back to its salt and preserving the inequality symbols, we obtain

$$\text{K}_3[\text{Fe(CN)}_6] \; \lt \; \text{K}_2\text{CrO}_4 \; \lt \; \text{KBr} = \text{KNO}_3 = \text{AlCl}_3.$$

This exact order is listed in Option A.

Hence, the correct answer is Option A.