A solution of m-chloroaniline, m-chlorophenol and m-chlorobenzoic acid in ethyl acetate was extracted initially with a saturated solution of NaHCO$$_3$$ to give fraction A. The left over organic phase was extracted with dilute NaOH solution to give fraction B. The final organic layer was labelled as fraction C. Fractions A, B and C, contain respectively:

We have a mixture that contains three compounds dissolved in the organic solvent ethyl acetate:

$$$\text{m-chloroaniline (basic amine)}$$$ $$$\text{m-chlorophenol (weak acid, }pK_a\approx 10\text{)}$$$ $$$\text{m-chlorobenzoic acid (stronger acid, }pK_a\approx 4\text{)}$$$

The separation is carried out by successive extractions, using the difference in acid-base strength of the functional groups. A compound converts to its water-soluble ionic form only if the reagent present is strong enough to protonate or deprotonate it. We trace each step.

First extraction: a saturated solution of $$\text{NaHCO}_3$$ (a weak base) is shaken with the organic layer.

The reaction possible is:

$$$\text{m-chlorobenzoic acid (HA)} + \text{HCO}_3^- \;\longrightarrow\; \text{m-chlorobenzoate ion (A}^-)\;+\;\text{H}_2\text{CO}_3$$$

Because $$pK_a(\text{benzoic acid})\approx 4$$ is sufficiently low, $$\text{HCO}_3^-$$ can remove its proton. Phenol, with $$pK_a\approx 10$$, is not acidic enough to be deprotonated by $$\text{HCO}_3^-$$, and the basic amine is certainly not affected by a base. Thus only the salt of m-chlorobenzoic acid transfers to the aqueous layer. We label this aqueous layer as Fraction A.

So, after the first extraction:

$$$\text{Fraction A (aqueous)}:\; \text{m-chlorobenzoate ion }(\;\longrightarrow\; \text{recovers m-chlorobenzoic acid on acidification)}$$$ $$$\text{Organic phase remaining}:\; \text{m-chlorophenol + m-chloroaniline}$$$

Second extraction: the residual organic layer is now shaken with dilute $$\text{NaOH}$$ (a stronger base than $$\text{NaHCO}_3$$).

The reaction that occurs is:

$$$\text{m-chlorophenol (ArOH)} + \text{OH}^- \;\longrightarrow\; \text{m-chlorophenoxide ion (ArO}^-)\;+\;\text{H}_2\text{O}$$$

This time the base is strong enough to deprotonate the phenol. The resulting phenoxide ion is water-soluble and goes into the new aqueous layer, which we label as Fraction B. The amine does not react with base, so it stays in the organic phase.

Thus after the second extraction:

$$$\text{Fraction B (aqueous)}:\; \text{m-chlorophenoxide ion }(\;\longrightarrow\; \text{recovers m-chlorophenol on acidification)}$$$ $$$\text{Organic phase left}:\; \text{m-chloroaniline}$$$

The final organic layer is labelled Fraction C, and it contains only the unreacted base.

Summarising the contents in their neutral, isolated forms obtained after the usual acidification of the aqueous fractions:

$$$\text{Fraction A} \;:\; \text{m-chlorobenzoic acid}$$$ $$$\text{Fraction B} \;:\; \text{m-chlorophenol}$$$ $$$\text{Fraction C} \;:\; \text{m-chloroaniline}$$$

Matching this order with the given options, we see that Option B lists the compounds in exactly this sequence.

Hence, the correct answer is Option B.