First, determine the molar mass of the compound C$_7$H$_5$N$_3$O$_6$:

Carbon (C): 12 g/mol, and there are 7 C atoms: $7 \times 12 = 84 \text{ g/mol}$

Hydrogen (H): 1 g/mol, and there are 5 H atoms: $5 \times 1 = 5 \text{ g/mol}$

Nitrogen (N): 14 g/mol, and there are 3 N atoms: $3 \times 14 = 42 \text{ g/mol}$

Oxygen (O): 16 g/mol, and there are 6 O atoms: $6 \times 16 = 96 \text{ g/mol}$

Calculate the total molar mass:

$ \text{Molar mass of C}_7\text{H}_5\text{N}_3\text{O}_6 = 84 + 5 + 42 + 96 = 227 \text{ g/mol} $

Calculate the number of moles of C$_7$H$_5$N$_3$O$_6$ in 681 grams:

$ \text{Moles of C}_7\text{H}_5\text{N}_3\text{O}_6 = \frac{681 \text{ g}}{227 \text{ g/mol}} = 3 \text{ moles} $

Each molecule of C$_7$H$_5$N$_3$O$_6$ contains 3 nitrogen atoms. Thus, in 3 moles, the total moles of nitrogen atoms is:

$ \text{Moles of N atoms} = 3 \times 3 = 9 \text{ moles} $

Using Avogadro's number, calculate the number of nitrogen atoms:

$ \text{Number of N atoms} = 9 \text{ moles} \times 6.02 \times 10^{23} \text{ atoms/mole} $

$ = 54.18 \times 10^{23} = 5.418 \times 10^{24} \text{ N atoms} $

Convert this number to the form $ x \times 10^{21} $:

$ 5.418 \times 10^{24} = 5418 \times 10^{21} $

Thus, the value of $ x $ is:

$ x = 5418 $