Step 1: Count carbon atoms:
Formic acid has $1$ carbon. The first higher homologue has $2$ carbons, the second has $3$ carbons.
Step 2: Name:
A carboxylic acid with three carbons is propanoic acid, $\text{C}_2\text{H}_5\text{COOH}$.
Butanoic acid has four carbons, so it is the third higher homologue, not the second.
Final Answer:
The answer is propanoic acid, option (C).
\[ \boxed{\text{Propanoic acid}} \]
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