In this reaction, the alkyl group \(\text{CH}_2\text{CH}_3\) will undergo radical substitution in the presence of UV light, and the major product will be a substitution of the hydrogen atom with a bromine atom. The resulting product is the alkyl bromide.
Thus, the correct answer is \( \boxed{\text{C}_6\text{H}_5\text{CH}_2\text{CH}_3\text{Br}} \).