C17 (18-3, Finland) It is known that a rectangular box can be filled with unit cubes. If it is instead filled with as many volume-2 cubes as possible, and these cubes are aligned parallel to 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 $(\sqrt[3]{2}=1.2599 \cdots)$.
Let the edge lengths of the rectangular box be positive integers $a_{1} \leqslant a_{2} \leqslant a_{3}$, and let $b_{i}=\left[\frac{a_{i}}{\sqrt[3]{2}}\right], i=$
$1,2,3$. According to the problem, we have
$$
\begin{aligned}
40 \% \cdot a_{1} a_{2} a_{3} & =2 b_{1} b_{2} b_{3}, \\
\frac{a_{1} a_{2} a_{3}}{b_{1} b_{2} b_{3}} & =5 .
\end{aligned}
$$

That is,
the values of $a, b, \frac{a}{b}$ are listed in the table below $\left(\right.$ where $\left.b=\left[\frac{a}{\sqrt[3]{2}}\right]\right)$ :
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
\hline$a$ & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
\hline$b$ & 1 & 2 & 3 & 3 & 4 & 5 & 6 & 7 & 7 \\
\hline$\frac{a}{b}$ & 2 & 1.5 & 1.3 & 1.6 & 1.5 & 1.4 & 1.3 & 1.23 & 1.42 \\
\hline
\end{tabular}

When $a>10$, $b \geqslant\left[\frac{11}{\sqrt[3]{2}}\right]=8$, so
$$
\begin{aligned}
\frac{a}{b}= & \frac{a}{\left[\frac{a}{\sqrt[3]{2}}\right]}2$, then $\frac{a_{1} a_{2} a_{3}}{b_{1} b_{2} b_{3}}<\left(\frac{5}{3}\right)^{3}<5$, which is impossible. Therefore, $a_{1}=2$.
If $a_{1}=a_{2}=2$, then by (1), $\frac{a_{3}}{b_{3}}=\frac{5}{4}<\sqrt[3]{2}$, which is impossible.
If $a_{1}=2, a_{2} \geqslant 6$, then $\frac{a_{2} a_{3}}{b_{2} b_{3}}=\left(\frac{3}{2}\right)^{2}<\frac{5}{2}$, contradicting (1). Hence, $3 \leqslant a_{2} \leqslant 5$. When $a_{1}=2, a_{2}=3$, then $\frac{a_{3}}{b_{3}}=\frac{5}{3}$, so $a_{3}=5$; when $a_{1}=2, a_{2}=4$, then $\frac{a_{3}}{b_{3}}=\frac{15}{8}$, which is impossible; when $a_{1}=2, a_{2}=5$, then $\frac{a_{3}}{b_{3}}=\frac{3}{2}$, which is only possible when $a_{3}=6$.
In summary, the dimensions of the box can only be $2 \times 3 \times 5$ and $2 \times 5 \times 6$.