Answer:

[tex]\qquad\boxed{\bf \: Volume_{CO_{2} \: in \: Air} = 1.8 \times {10}^{5} \: {cm}^{3} }[/tex]

[tex]\Large\underline{\mathbb{ SOLUTION }}\longrightarrow[/tex]

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Given that

[tex]\sf \: Volume_{Air} = 6 \times {10}^{8} \: {cm}^{3} [/tex]

And

[tex]\sf \: Volume_{CO_{2} \: in \: Air} = 0.03\%[/tex]

So Volume of Carbon Dioxide =

[tex]\sf \: 0.03\% \: \: of \: \: 6 \times {10}^{8} \\ \\ \sf \: = \frac{0.03}{100} \times 6 \times {10}^{8} \\ \\ \sf \: \frac{3}{10000 } \times 6 \times {10}^{8} \\ \\ \sf \: \frac{18}{ {10}^{4} } \times {10}^{8} \\ \\ \sf \: 18 \times {10}^{8 - 4} \\ \\ \boxed{\sf \: \because \: \frac{ {x}^{m} }{ {x}^{n} } = {x}^{m - n} } \\ \\ \sf \: 18 \times {10}^{4} \\ \\ \implies \: \sf \: 1.8 \times {10}^{5} \: {cm}^{3} \\ \\ \implies \: \boxed{\bf \:Volume_{CO_{2}\: in \: Air} = 1.8 \times {10}^{5} \: {cm}^{3} }[/tex]

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