From 200 mg of  ${\text{CO}}_2,{10}^{21}$ molecules are removed. Then the mass and number of moles of  ${\text{CO}}_2$ left are

$\begin{array}{l}44{\text{\hspace{0.17em}g\hspace{0.17em}of\hspace{0.17em}CO}}_2\to 6.023\times {10}^{23}\text{molecules}\\ 200\times {10}^{-3}g{\text{ of CO}}_2\to ?\\ \frac{200\times 6.023\times {10}^{23}\times {10}^{-3}}{44}\\ =2.737\times {10}^{21}{\text{molecules\hspace{0.17em}of\hspace{0.17em}CO}}_2\end{array}$

  From   $2.737\times {10}^{21}$molecules of ${\text{CO}}_{2}1\times {10}^{21}$molecules are removed

$\frac{\begin{array}{l}2.737\times {10}^{21}\\ -1.000\times {10}^{21}\end{array}}{1.737\times {10}^{21}}$

Molecules of   ${\text{CO}}_2$  are left 

$\begin{array}{l}i)6.023\times {10}^{23}\to 44\text{g}\\ 1.737\times {10}^{21}\to ?\\ =\frac{1.737\times 44\times {10}^{21}}{6.023\times {10}^{23}}\\ =12.68\times {10}^{-2}\\ =126.8\times {10}^{-3}\text{g\hspace{0.17em}\hspace{0.17em}}of{\text{\hspace{0.17em}\hspace{0.17em}CO}}_{2}left\end{array}$

$\begin{array}{l}ii)1\text{\hspace{0.17em}mole}\to 6.023\times {10}^{\text{23}}\text{molecules}\\ ?\to 1.737\times {10}^{21}\text{molecules}\\ =\frac{1.737\times {10}^{21}}{6.023\times {10}^{23}}=\frac{1.737\times {10}^{-2}}{6.023}\\ =0.288\times {10}^{-2}\\ =2.88\times {10}^{-3}{\text{moles\hspace{0.17em}of\hspace{0.17em}CO}}_2\text{ left}\end{array}$