This is a numerical-answer question. The correct answer is shown below.
[Solution by Dr. Abubacker] $\text{C}_6\text{H}_5\text{COOH}+\text{NaHCO}_3\to\text{C}_6\text{H}_5\text{COONa}+\text{H}_2\text{O}+\text{CO}_2$. $11.2$ L at STP $=0.5$ mol $\text{CO}_2$, so $0.5$ mol of acid is needed: $0.5\times122=61$ g. Answer: 61.