Step 1 - Recall the chemical identity of aspirin.
Aspirin is ACETYLSALICYLIC ACID, i.e. 2-acetoxybenzoic acid.

Step 2 - Work backwards from the product.
Removing the acetyl group $(CH_3CO-)$ from the ester oxygen of aspirin regenerates the
phenolic $-OH$. The parent compound is therefore SALICYLIC ACID, whose IUPAC name is
2-HYDROXYBENZOIC ACID.

Step 3 - Write the reaction.
$$\text{2-Hydroxybenzoic acid} + (CH_3CO)_2O \xrightarrow{\ H^+\ } \text{Aspirin} + CH_3COOH$$
The phenolic $-OH$ at position 2 is acetylated to $-OCOCH_3$, while the $-COOH$ group
remains unchanged.

Step 4 - Reject the other options.
Phenol has no $-COOH$ group; salicylaldehyde has $-CHO$ instead of $-COOH$;
benzoic acid has no phenolic $-OH$ to be acetylated.

Answer: (C) 2-Hydroxybenzoic acid