Step 1: Spot the pattern.
An alkyl halide plus the silver salt of an acid gives an ester plus silver halide.
Step 2: Assemble the ester.
Alkyl group from the halide: $\text{C}_2\text{H}_5$. Acyl part from the silver salt: $\text{CH}_3\text{CO}$. Joined through oxygen, we get $\text{CH}_3\text{COOC}_2\text{H}_5$.
Step 3: Name and compare.
This is ethyl acetate, option (C). Methyl acetate (D) would contain a methyl group from the halide, which is not our case.
Final Answer:
Option (C).
\[ \boxed{\text{CH}_3\text{COOC}_2\text{H}_5} \]