We name a,b,c the sides of the parallelepiped, which are positive integers. We also put \[ x = \left\lfloor\frac{a}{\sqrt[3]{2}}\right\rfloor \ \ \ \ y = \left\lfloor\frac{b}{\sqrt[3]{2}}\right\rfloor \ \ \ \ z = \left\lfloor\frac{c}{\sqrt[3]{2}}\right\rfloor \ \ \ \] It is clear that $ xyz$ is the maximal number of cubes with sides of length $ \sqrt[3]{2}$ that can be put into the parallelepiped with sides parallels to the sides of the box. Hence the corresponding volume is $ V_2=2\cdot xyz$. We need $ V_2=0.4\cdot V_1=0.4\cdot abc$, hence \[ \frac ax\cdot \frac by\cdot \frac cz=5\ \ \ \ \ \ \ \ (1)\] We give the values of $ x$ and $ a/x$ for $ a=1,\dots ,8$. The same table is valid for $ b,y$ and $ c,z$. \[ \begin{tabular}{|c|c|c|} \hline a & x & a/x \\ \hline 1 & 0 & - \\ \hline 2 & 1 & 2 \\ \hline 3 & 2 & 3/2 \\ \hline 4 & 3 & 4/3 \\ \hline 5 & 3 & 5/3 \\ \hline 6 & 4 & 3/2 \\ \hline 7 & 5 & 7/5 \\ \hline 8 & 6 & 4/3 \\ \hline \end{tabular}\] By simple inspection we obtain two solutions of $ (1)$: $ \{a,b,c\}=\{2,5,3\}$ and $ \{a,b,c\}=\{2,5,6\}$. We now show that they are the only solutions. We can assume $ \frac ax\ge \frac by \ge \frac cz$. So necessarily $ \frac ax\ge \sqrt[3]{5}$. Note that the definition of $ x$ implies \[ x< a/\sqrt[3]2 < x+1,\] hence \[ \sqrt[3]2< a/x < \sqrt[3]2(1+\frac 1x)\] If $ a\ge 4$ then $ x\ge 3$ and $ \frac ax<\sqrt[3]2(1+\frac 1x)\le \sqrt[3]2(\frac 43)<\sqrt[3]5$ since $ 2\cdot \frac {4^3}{3^3}<5$. So we have only left the cases $ a=2$ and $ a=3$. But for $ a=3$ we have $ a/x=3/2<\sqrt[3]5$ and so necessarily $ a=2$ and $ a/x=2$. It follows \[ \frac by \cdot \frac cz =\frac 52 \ \ \ \ \ \ (2)\] Note that the definitions of $ y,z$ imply \[ y< b/\sqrt[3]2 < y+1,\ \ \textrm{and} \ \ z< c/\sqrt[3]2 < z+1.\ \ \ \ (3)\] Moreover we have from (2) and from $ b/y\ge c/z$ that \[ \frac by \ge \sqrt{5/2}\ \ \ \ \ (4)\] If $ b=2$ then $ b/y=2$ and we would have $ c/z=5/4<\sqrt[3]2$, which contradicts $ (3)$. On the other hand, if $ b>5$ then $ y>4$ and $ \frac by<\sqrt[3]2(1+\frac 1y)\le \sqrt[3]2(\frac 54)<\sqrt{5/2}$ since $ 2^2\cdot \frac {5^6}{4^6}<\frac{5^3}{2^3}$ as $ 5^3<2^7$. So we have only left the cases $ b=3,4,5$. But for $ b=3$ we have $ b/y=3/2<\sqrt{5/2}$ and for $ b=4$ we have $ b/y=4/3<\sqrt{5/2}$ and so necessarily $ b=5$ and $ b/y=5/3$ ($ >\sqrt{5/2}$) So we arrive finally at $ a=2,b=5$ and $ c/z=3/2$. If $ c\ge 8$ then $ z\ge 6$ and $ \frac cz<\sqrt[3]2(1+\frac 1z)\le \sqrt[3]2(\frac 76)<\frac 32$ since $ 2\cdot \frac {7^3}{6^3}<\frac{3^3}{2^3}$. On the other hand, for $ c\le 7$ there are the only two possible values $ c=3$ and $ c=6$ which yield the known solutions.

How can you fill a parallelpiped completely with cubes of dimension 1? Isn't this only possible when the faces are all either parallel or perpendicular? (i.e. a rectangular prism). Not sure why the wording includes "parallelpiped" rather than "rectangular prism." Am I misunderstanding something here?

The official wording is a rectangular box.