Step 1: Read the clues about A.
The compound A is $C_4H_9Br$. It reacts with sodium in dry ether (a Wurtz reaction) to give B. Then light-driven chlorination of B gives only two monochlorides. These clues fix what A is.
Step 2: Identify B from the Wurtz reaction.
The Wurtz reaction joins two alkyl halides into a bigger alkane. The bromide that gives an alkane producing just two monochlorides is the straight-chain $n$-butyl bromide. So B is $n$-octane.
Step 3: Confirm A is a primary bromide.
Since B comes from a straight-chain bromide, A is $n$-butyl bromide, a primary alkyl halide ($CH_3CH_2CH_2CH_2Br$).
Step 4: Test the reaction with sodium ethoxide.
Primary alkyl halides favour substitution over elimination with a nucleophile. With sodium ethoxide, \[ RBr + NaOC_2H_5 \rightarrow ROC_2H_5 + NaBr, \] so substitution mainly happens. This matches the correct option.
Step 5: Reject the other statements.
A primary bromide has no four different groups on one carbon, so it is not chiral. Primary halides do not go by $S_N1$ (they prefer $S_N2$). Adding HBr to but-2-ene gives 2-bromobutane, not $n$-butyl bromide.
Step 6: State the correct statement.
So the right statement about A is \[ \boxed{\text{reaction with } NaOC_2H_5 \text{ gives mainly substitution}} \]