A box whose shape is a parallelepiped can be completely filled with cubes of side $1.$ If we put in it the maximum possible number of cubes, each of volume $2$, with the sides parallel to those of the box, then exactly $40$ percent of the volume of the box is occupied. Determine the possible dimensions of the box.
1. **Define the variables and initial conditions:**
   Let \(a, b, c\) be the sides of the parallelepiped, which are positive integers. We are given that the box can be completely filled with unit cubes, so the volume of the box is \(V_1 = abc\).

2. **Determine the maximum number of cubes with volume 2:**
   Let \(x, y, z\) be the maximum number of cubes with side length \(\sqrt[3]{2}\) that can fit along the sides \(a, b, c\) respectively. Thus,
   \[
   x = \left\lfloor \frac{a}{\sqrt[3]{2}} \right\rfloor, \quad y = \left\lfloor \frac{b}{\sqrt[3]{2}} \right\rfloor, \quad z = \left\lfloor \frac{c}{\sqrt[3]{2}} \right\rfloor
   \]
   The volume occupied by these cubes is \(V_2 = 2 \cdot xyz\).

3. **Relate the volumes:**
   We are given that \(V_2 = 0.4 \cdot V_1\). Therefore,
   \[
   2 \cdot xyz = 0.4 \cdot abc
   \]
   Simplifying, we get:
   \[
   5 \cdot xyz = abc \quad \text{or} \quad \frac{a}{x} \cdot \frac{b}{y} \cdot \frac{c}{z} = 5
   \]

4. **Analyze the possible values of \(a, b, c\):**
   We create a table for \(a\) and \(x\) values:
   \[
   \begin{array}{|c|c|c|}
   \hline
   a & x & \frac{a}{x} \\
   \hline
   1 & 0 & - \\
   2 & 1 & 2 \\
   3 & 2 & \frac{3}{2} \\
   4 & 3 & \frac{4}{3} \\
   5 & 3 & \frac{5}{3} \\
   6 & 4 & \frac{3}{2} \\
   7 & 5 & \frac{7}{5} \\
   8 & 6 & \frac{4}{3} \\
   \hline
   \end{array}
   \]

5. **Identify possible solutions:**
   By inspection, we find that the values \(a = 2, b = 5, c = 3\) and \(a = 2, b = 5, c = 6\) satisfy the equation \(\frac{a}{x} \cdot \frac{b}{y} \cdot \frac{c}{z} = 5\).

6. **Verify the solutions:**
   - For \(a = 2, b = 5, c = 3\):
     \[
     x = 1, \quad y = 3, \quad z = 2 \quad \Rightarrow \quad \frac{2}{1} \cdot \frac{5}{3} \cdot \frac{3}{2} = 5
     \]
   - For \(a = 2, b = 5, c = 6\):
     \[
     x = 1, \quad y = 3, \quad z = 4 \quad \Rightarrow \quad \frac{2}{1} \cdot \frac{5}{3} \cdot \frac{6}{4} = 5
     \]

7. **Show these are the only solutions:**
   Assume \(\frac{a}{x} \geq \frac{b}{y} \geq \frac{c}{z}\). Then \(\frac{a}{x} \geq \sqrt[3]{5}\). Given \(x < \frac{a}{\sqrt[3]{2}} < x + 1\), we have:
   \[
   \sqrt[3]{2} < \frac{a}{x} < \sqrt[3]{2} \left(1 + \frac{1}{x}\right)
   \]
   For \(a \geq 4\), \(x \geq 3\), and \(\frac{a}{x} < \sqrt[3]{2} \left(\frac{4}{3}\right) < \sqrt[3]{5}\). Thus, \(a = 2\) or \(a = 3\). For \(a = 3\), \(\frac{a}{x} = \frac{3}{2} < \sqrt[3]{5}\), so \(a = 2\).

8. **Determine \(b\) and \(c\):**
   From \(\frac{b}{y} \cdot \frac{c}{z} = \frac{5}{2}\) and \(\frac{b}{y} \geq \sqrt{\frac{5}{2}}\), we find \(b = 5\) and \(\frac{b}{y} = \frac{5}{3}\). For \(c\), we find \(c = 3\) or \(c = 6\).

The final answer is \( \boxed{ \{a, b, c\} = \{2, 5, 3\} } \) or \(\{2, 5, 6\}\).