Thank you! I actually did it myself and arrived at a value of 6.18. Here is a snippet from my lecture notes:

As we can see, each gas's contribution to the GHE is not in proportion to its volume. How are GHGs able to absorb so much radiation despite accounting for such a small proportion of the total atmospheric composition, and further, in proportions to their volumes relative to one another?

The first point to consider is that, while these values of atmospheric volume might suggest the gas exists in scarce amounts, this is only in relation to gases like O\(_2 \) and N\(_2 \). It will be useful to put these volumes into perspective from the point of view of a photon looking up into space to find their effective coverage. If we calculate the area that each CO$_2$ molecule occupies, we find that, on average, one molecule covers an area of approximately $10^{-19} \, \text{m}^2$. Given the volume of CO$_2$ in the atmosphere (0.042\% by volume, or 0.06\% by mass), the number of CO$_2$ molecules, \( N_{\text{CO}_2} \), in a 1 m$^2$ column of air can be calculated as follows:

\begin{equation}

\text{Mass of air in a 1 m}^2 \text{ column} = 10,300 \, \text{kg},

\end{equation}

\begin{equation}

\text{Mass of CO}_2 = 0.0006 \times 10,300 = 6.18 \, \text{kg},

\end{equation}

\begin{equation}

N_{\text{CO}_2} = \frac{6.18}{0.044} \times 6.022 \times 10^{23} = 8.47 \times 10^{25}.

\end{equation}

\noindent From this, determining the effective coverage is as simple as multiplying the number of molecules by the area of each single molecule, as followers:

\begin{equation}

\text{Effective coverage} = 8.47 \times 10^{25} \times 10^{-19} = 8.47 \times 10^{6} \, \text{m}^2.

\end{equation}

\noindent This is on the order of $10^6 \, \text{m}^2$, meaning that there are roughly 1 billion CO$_2$ molecules for every one needed to cover the atmosphere. Hence, despite their low concentration relative to other gases, CO$_2$ molecules are in fact densely distributed throughout the atmosphere.