Identify the two substrates

• Benzyl bromide: $\mathrm{C_6H_5CH_2Br}$
• Ethyl bromide: $\mathrm{CH_3CH_2Br}$

We need to compare their reactivity in the $\mathrm{S_N2}$ reaction.

Check Assertion (A)

In an $\mathrm{S_N2}$ reaction, the nucleophile attacks from the back side and passes through a transition state in which the carbon is approximately pentacoordinate (trigonal bipyramidal-like).

Benzyl halides are known to undergo $\mathrm{S_N2}$ reactions faster than normal primary alkyl halides such as ethyl bromide.

Therefore, $\mathrm{C_6H_5CH_2Br}$ reacts more readily by $\mathrm{S_N2}$ than $\mathrm{CH_3CH_2Br}.$

So, Assertion (A) is correct.

Check Reason (R)

In the $\mathrm{S_N2}$ transition state for benzyl bromide, the reacting benzylic carbon develops partial bonding to both the incoming nucleophile and the leaving group. This transition state has increased p-character / a partially developed unhybridized $p$-orbital character that can overlap with the adjacent phenyl ring.

Because the benzylic center is directly attached to the benzene ring, this transition state is resonance stabilized by conjugation with the phenyl ring.

This lowers the activation energy and makes the $\mathrm{S_N2}$ reaction faster.

Hence, Reason (R) is also correct.

Does (R) explain (A)?

Yes. The enhanced $\mathrm{S_N2}$ reactivity of benzyl bromide compared to ethyl bromide is due to stabilization of the transition state by conjugation with the phenyl ring.

Thus, (R) is the correct explanation of (A).

Correct option

Therefore, the correct answer is: $\boxed{\text{B}}$

Both (A) and (R) are correct and (R) is the correct explanation of (A).