The sequence of reactions starts with the nitration of benzene to form nitrobenzene (B), which undergoes further reactions leading to the formation of compound E. The atomic mass of nitrogen is 14, and the molecular weight of compound E is given by the sum of the atomic masses of all the elements involved.
\[
\text{Molecular mass of compound E} = \text{mass of C}_6\text{H}_6 + 1 \times \text{mass of N} + 3 \times \text{mass of O}
\]
\[
= 78 + 14 + 48 = 140 \, \text{g/mol}
\]
Thus, the percentage of nitrogen in compound E is:
\[
\frac{\text{mass of N}}{\text{molecular mass of E}} \times 100 = \frac{14}{140} \times 100 = 10%
\]
So, the percentage of nitrogen in compound E is \( \boxed{20} \).