Use combustion data to find mass of carbon

From combustion, all carbon in the compound appears in $\mathrm{CO_2}$.

Given: $m(\mathrm{CO_2}) = 0.20\,\text{g}$

In $44\,$g of $\mathrm{CO_2}$, carbon mass is $12\,$g.

Therefore, carbon in $0.20\,$g of $\mathrm{CO_2}$ is $m_C = 0.20\times \frac{12}{44}$ $m_C = 0.20\times \frac{3}{11} = \frac{0.60}{11} \approx 0.0545\,\text{g}$

Find percentage of carbon in the original compound

Mass of organic compound taken: $m_{\text{compound}} = 0.30\,\text{g}$

So, $\%\,C = \frac{0.0545}{0.30}\times 100$ $\%\,C \approx 18.18\%$

Nearest integer

$\boxed{18}$

Comparison with stored answer

Stored correct answer = $18$

Our derived answer matches it.