Answer
(c) 2.0 L
Work Step by Step
The total volume of the gas-exchanging region of the lungs is computed by multiplying the total number of alveoli by the average volume of a single alveolus:
$$
\begin{align}
V &= \left( \textrm{total number of alveoli} \right) \times \left( \textrm{average volume of a single alveolus} \right) \\
&= \left( 480 \times 10^6 \right) \times \left( 4.2 \times 10^6 \,\mu\textrm{m}^3 \right) \\
&= 2016 \times 10^{12} \,\mu\textrm{m}^3
\end{align}
$$ We want to convert $\mu\textrm{m}^3$ to $\textrm{L}$, so we can use the fact that $$ 1 \,\textrm{cm}^3 = 1 \,\textrm{mL} $$keeping in mind that $\textrm{cm}^3$ is cubic centimeters. The process of converting $\mu\textrm{m}^3$ to $\textrm{L}$ is as follows:
$$ \mu\textrm{m}^3 \longrightarrow \textrm{cm}^3 \longrightarrow \textrm{mL} \longrightarrow \textrm{L}
$$ Now, we have the following:
$$
2016 \times 10^{12} \,\mu\textrm{m}^3 \times \left( \dfrac{10^{-6} \,\textrm{m}}{1 \,\mu\textrm{m}} \times \dfrac{1 \,\textrm{cm}}{10^{-2} \,\textrm{m}} \right)^3 \times \left( \dfrac{1 \,\textrm{mL}}{1 \,\textrm{cm}^3} \right) \times \left( \dfrac{10^{-3} \,\textrm{L}}{1 \,\textrm{mL}} \right)
$$ which simplifies to $2.016 \,\textrm{L}$, which is rounded off to $2.0 \,\textrm{L}$. Thus, the correct answer is (c).
