$( N _{ A }=6.02 \times 10^{23} \,mol ^{-1})$(નજીકનો પૂર્ણાંક)
\(n _{ C _{7} H _{5} N _{3} O _{6}}=\frac{681}{227}=3\)
\(n _{ N }=\frac{681}{227} \times 3=9 mol\)
no. of \(N\) atoms \(=9 \times 6.02 \times 10^{23}\)
\(=5418 \times 10^{21}\)
\(\therefore\) The answer is \(5418 .\)