$( N _{ A }=6.02 \times 10^{23} \,mol ^{-1})$(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 $5418 .$
[આણ્વિય વજન : ${H}=1.008 ; {C}=12.00 ; {O}=16.00$ ]