(B) The $E. coli$ genome consists of $4.6 \times 10^6$ base pairs.The distance between two consecutive base pairs is $0.34 \times 10^{-9} \; m$ or $0.

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(B) The $E. coli$ genome consists of $4.6 \times 10^6$ base pairs.The distance between two consecutive base pairs is $0.34 \times 10^{-9} \; m$ or $0.34 \times 10^{-6} \; mm$.Therefore,the total length of $DNA$ is calculated as: $4.6 \times 10^6 \times 0.34 \times 10^{-6} \; mm = 1.36 \; mm$.