We know that $\mathrm{S_N2}$ mechanism involves a transition state wherein both, the incoming nucleophile as well as the leaving group are present around the carbon atom. There are 5 atoms simultaneously bonded to it. Thus, for such a transition state to be possible,there should be minimum steric hindrance. Hence, 1° alkyl halides are most reactive towards $\mathrm{S_N2}$ followed by 2° and finally 3°. 1° RX > 2° RX > 3° RX Based on the above order, (i) $\mathrm{CH_3CH_2CH_2CH_2Br}$ is more reactive. (ii) $\mathrm{CH}_3\mathrm{CH}_2\underset{\vert_{\mathrm{Br}}\mathrm{C}}{\mathrm{HCH}_3}$ is more reactive. (iii) $\mathrm{CH}_3-\underset{\vert_{\mathrm{CH}_3}\mathrm{C}}{\mathrm{H-CH}_2}-\mathrm{CH}_2-\mathrm{Br}$ is more reactive. In (iii), it is the proximity of the branched chain $\mathrm{—CH}_3$ that determines the reactivity. In $\mathrm{CH_3CH_2CH(CH_3)CH_2}$ the methyl group is closer to the leaving group thereby hindering the transition state.