The increasing order of the acidity of the following carboxylic acids is:

The acidic strength of a carboxylic acid depends on how stable its conjugate base (carboxylate anion) is. Greater stability of the anion means stronger acidity because the equilibrium $$\ce{RCOOH \rightleftharpoons RCOO^- + H^+}$$ lies further to the right.

Stability of the anion is increased by electron-withdrawing groups (EWG, $$-I$$ effect) and decreased by electron-donating groups (EDG, $$+I$$ or $$+R$$ effects). Therefore:

• A strong EWG (-NO2) makes the anion most stable → strongest acid.
• A halogen (-Cl) has a strong $$-I$$ effect but a weak $$+R$$ effect; net result: more acidic than the unsubstituted compound.
• No substituent (-H) is taken as the reference, i.e. benzoic acid itself.
• A methyl group (-CH3) is an EDG due to its $$+I$$ effect; it destabilises the anion → weakest acid.

Assigning the given acids:
  I : p-nitro­benzoic acid  (strong EWG)
  II: benzoic acid  (no substituent)
  III: p-methyl­benzoic acid  (EDG)
  IV: p-chloro­benzoic acid  (halogen EWG)

Accordingly, the stability of their conjugate bases (and hence their acidity) follows the order

$$\text{p-Me-C}_6\text{H}_4\text{COOH} \; \lt \; \text{C}_6\text{H}_5\text{COOH} \; \lt \; \text{p-Cl-C}_6\text{H}_4\text{COOH} \; \lt \; \text{p-NO}_2\text{-C}_6\text{H}_4\text{COOH}$$

That is, $$\text{III} \lt \text{II} \lt \text{IV} \lt \text{I}$$.

Hence the correct increasing order of acidity is Option A: III < II < IV < I.