We know that SN2 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 SN2 followed by 2° and finally 3°.
1° RX > 2° RX > 3° RX
Based on the above order,
(i) CH3CH2CH2CH2Br is more reactive.
(ii) CH3CH2

C

| Br

HCH3 is more reactive.
(iii) CH3−

C

| CH3

H−CH2−CH2−Br is more reactive.
In (iii), it is the proximity of the branched chain —CH3 that determines the reactivity. In CH3CH2CH(CH3)CH2 the methyl group is closer to the leaving group thereby hindering the transition state.