Step 1: Draw and mark:
Chain: C1($\text{H}_3$)-C2(H)(I)-C3(H)($\text{CH}_3$)-C4(H)($\text{CH}_3$)-C5(H)($\text{CH}_3$)-C6($\text{H}_3$).
Step 2: Look for repeated groups:
C5 carries a methyl branch and also the terminal C6 methyl, which are identical, so C5 is achiral. C2, C3 and C4 each have four different groups. Total = 3 (B).