4-213 Given a rectangular box that can be filled with unit cubes. If it is instead filled with as many volume-2 cubes as possible, and these cubes are aligned with the edges of the box, then the box's volume is exactly filled to $40 \%$. Try to find the volume of a rectangular box with this property $(\sqrt[3]{2}=1.2599 \cdots)$.
[Solution] Let $Q$ be a rectangular box with the required properties, and let $a, b, c (a \leqslant b \leqslant c)$ be the side lengths of $Q$ that meet the conditions. Since this rectangular box can be filled with unit cubes, $a, b, c$ are natural numbers. Now, placing a cube with a volume of 2, its edge length is $\sqrt[3]{2}$, so we know $a \geqslant 2$.

We know that on a line segment of length $n$, using a line segment of length $\sqrt[3]{2}$ to measure, it can only measure $\left[\frac{n}{\sqrt[3]{2}}\right]$ times, and the remaining part is less than $\sqrt[3]{2}$. Here $\left[\frac{n}{\sqrt[3]{2}}\right]$ represents the largest integer not exceeding $\frac{n}{\sqrt[3]{2}}$. Therefore, the box can accommodate
$$
\left[\frac{a}{\sqrt[3]{2}}\right]\left[\frac{b}{\sqrt[3]{2}}\right]\left[\frac{c}{\sqrt[3]{2}}\right]
$$

cubes with a volume of 2. Since the volume is 40% of $a b c$,
we get $2 \cdot\left[\frac{a}{\sqrt[3]{2}}\right]\left[\frac{b}{\sqrt[3]{2}}\right]\left[\frac{c}{\sqrt[3]{2}}\right]=0.4 a b c$.
Thus, $\frac{a}{\left[\frac{a}{\sqrt[3]{2}}\right]} \cdot \frac{b}{\left[\frac{b}{\sqrt[3]{2}}\right]} \cdot \frac{c}{\left[\frac{c}{\sqrt[3]{2}}\right]}=5$.

To find the natural numbers $a, b, c$ that satisfy the above equation, we can first consider the function
$$
g(n)=\frac{n}{\left[\frac{n}{\sqrt[3]{2}}\right]} \quad(n \geqslant 2) .
$$

The table is as follows:
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|}
\hline$n$ & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & $\cdots$ \\
\hline$\left[\frac{n}{\sqrt[3]{2}}\right]$ & 1 & 2 & 3 & 3 & 4 & 5 & 6 & 7 & 7 & $\cdots$ \\
\hline$g(n)$ & $\frac{2}{1}$ & $\frac{3}{2}$ & $\frac{4}{3}$ & $\frac{5}{3}$ & $\frac{6}{4}$ & $\frac{7}{5}$ & $\frac{8}{6}$ & $\frac{9}{7}$ & $\frac{10}{7}$ & $\cdots$ \\
\hline
\end{tabular}

The function values $g(n)$ for $n \leqslant 10$ in the table above are obtained through specific calculations. For $n>10$, the estimation can be done as follows:
$$
\begin{aligned}
\frac{\left[\frac{n}{\sqrt[3]{2}}\right]}{n} & =\frac{\frac{n}{\sqrt[3]{2}}-\left\{\frac{n}{\sqrt[3]{2}}\right\}}{n}\left(\text { where }\left\{\frac{n}{\sqrt[3]{2}}\right\} \text { represents the fractional part of } \frac{n}{\sqrt[3]{2}}\right) . \\
& =\frac{1}{\sqrt[3]{2}}-\frac{1}{n}\left\{\frac{n}{\sqrt[3]{2}}\right\} \\
& >0.79-\frac{1}{n}\left\{\frac{n}{\sqrt[3]{2}}\right\} \\
& >0.79-\frac{1}{10} \\
& =0.69>\frac{2}{3} .
\end{aligned}
$$

So $\frac{n}{\left[\frac{n}{\sqrt[3]{2}}\right]}$ for $n>10$ is greater than 2, then
$$
g(a) g(b) g(c) \leqslant\left(\frac{5}{3}\right)^{3}=\frac{125}{27}<5.
$$
If $g(a) \leqslant 2$, then
$$
g(b) g(c) \geqslant \frac{5}{2}>\left(\frac{5}{3}\right)^{2}=\frac{25}{9},
$$
which implies $g(b) \geqslant \frac{5}{3}$ and $g(c) \geqslant \frac{5}{3}$. If $g(b) \geqslant 2$ and $g(c) \geqslant 2$, then
$$
g(a) g(b) g(c) \leqslant\left(\frac{5}{3}\right)^{3}=\frac{125}{27}<5.
$$
If $g(b) \geqslant 2$ and $g(c) \geqslant \frac{5}{4}$, then
$$
g(a) g(b) g(c) \leqslant 2 \cdot \frac{5}{3} \cdot \frac{5}{4}=\frac{25}{6}<5.
$$
If $g(b) \geqslant \frac{5}{3}$ and $g(c) \geqslant \frac{5}{4}$, then
$$
g(a) g(b) g(c) \leqslant \frac{5}{3} \cdot \frac{5}{3} \cdot \frac{5}{4}=\frac{125}{36}<5.
$$
If $g(b) \geqslant \frac{5}{3}$ and $g(c) \geqslant \frac{3}{2}$, then
$$
g(a) g(b) g(c) \leqslant \frac{5}{3} \cdot \frac{5}{3} \cdot \frac{3}{2}=\frac{25}{6}<5.
$$
If $g(b) \geqslant \frac{3}{2}$ and $g(c) \geqslant \frac{5}{3}$, then
$$
g(a) g(b) g(c) \leqslant \frac{5}{3} \cdot \frac{3}{2} \cdot \frac{5}{3}=\frac{25}{6}<5.
$$
If $g(b) \geqslant \frac{3}{2}$ and $g(c) \geqslant \frac{3}{2}$, then
$$
g(a) g(b) g(c) \leqslant \frac{5}{3} \cdot \frac{3}{2} \cdot \frac{3}{2}=\frac{15}{4}<5.
$$
Therefore, $g(b), g(c)$ are not less than $\frac{3}{2}$. After verification, we have
$$
g(b)=\frac{5}{3}, g(c)=\frac{3}{2} \text{ (or } g(b)=\frac{3}{2}, g(c)=\frac{5}{3} \text{. )}
$$

Thus, we have
$$
b=5, c=3 \text{ or } 6 \text{ (or } b=3 \text{ or } 6, c=5 \text{. )}
$$

Therefore, the possible internal dimensions of the box $Q$ are $(2,5,6)$ or $(2,3,5)$, and its volume is 60 or 30.