$$2.8 \times 10^{-3}$$ mol of $$CO_{2}$$ is left after removing $$10^{21}$$ molecules from its 'x' mg sample. The mass of $$CO_{2}$$ taken initially is Given: $$N_{A} = 6.02 \times 10^{23} mol^{-1}$$

We need to find the initial mass of $$CO_2$$.

After removing $$10^{21}$$ molecules from the sample, $$2.8 \times 10^{-3}$$ mol of $$CO_2$$ remains.

The moles of $$CO_2$$ removed can be calculated as $$ n_{removed} = \frac{10^{21}}{6.02 \times 10^{23}} = \frac{1}{602} = 1.66 \times 10^{-3} \text{ mol} $$.

Adding this to the remaining moles gives the total initial moles: $$ n_{initial} = 2.8 \times 10^{-3} + 1.66 \times 10^{-3} = 4.46 \times 10^{-3} \text{ mol} $$.

The molar mass of $$CO_2$$ is $$12 + 2(16) = 44$$ g/mol.

Therefore, the initial mass is $$ x = 4.46 \times 10^{-3} \times 44 = 0.19624 \text{ g} = 196.2 \text{ mg} $$.

The correct answer is Option 3: 196.2 mg.