$44\,gm$ of $C{O_2}$ = $6 \times {10^{23}}$ molecules

$0.2\,gm$ of  $C{O_2}$ = $\frac{{6 \times {{10}^{23}}}}{{44}} \times 0.2 = 0.0272 \times {10^{23}}$ $ = 2.72 \times {10^{21}}$ molecule

Now ${10^{21}}$ molecule are removed.

So remaining molecules $ = 2.72 \times {10^{21}} - {10^{21}}$ $ = {10^{21}}(2.72 - 1)$ =$1.72 \times {10^{21}}$ molecules

Now, $6.023 \times {10^{23}}$molecules = $1$ $mole$

$1.72 \times {10^{21}}$ molecules $ = \frac{{1 \times 1.72 \times {{10}^{21}}}}{{6.023 \times {{10}^{23}}}} = 0.285 \times {10^{ - 2}}$

$ = 2.85 \times {10^{ - 3}}$.