6. Suppose $a_{1} \leq a_{2} \leq a_{3}$ are the dimensions of the box. If we set $b_{i} = \left[a_{i} / \sqrt[3]{2}\right]$, the condition of the problem is equivalent to $\frac{a_{1}}{b_{1}} \cdot \frac{a_{2}}{b_{2}} \cdot \frac{a_{3}}{b_{3}} = 5$. We list some values of $a, b = [a / \sqrt[3]{2}]$ and $a / b$:

| $a$ | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $b$ | 1 | 2 | 3 | 3 | 4 | 5 | 6 | 7 | 7 |
| $a / b$ | 2 | 1.5 | 1.33 | 1.67 | 1.5 | 1.4 | 1.33 | 1.29 | 1.43 |

We note that if $a > 2$, then $a / b \leq 5 / 3$, and if $a > 5$, then $a / b \leq 3 / 2$. If $a_{1} > 2$, then $\frac{a_{1}}{b_{1}} \cdot \frac{a_{2}}{b_{2}} \cdot \frac{a_{3}}{b_{3}} < (5 / 3)^{3} < 5$, a contradiction. Hence $a_{1} = 2$. If also $a_{2} = 2$, then $a_{3} / b_{3} = 5 / 4 \leq \sqrt[3]{2}$, which is impossible. Also, if $a_{2} \geq 6$, then $\frac{a_{2}}{b_{2}} \cdot \frac{a_{3}}{b_{3}} \leq (1.5)^{2} < 2.5$, again a contradiction. We thus have the following cases:
(i) $a_{1} = 2, a_{2} = 3$, then $a_{3} / b_{3} = 5 / 3$, which holds only if $a_{3} = 5$;
(ii) $a_{1} = 2, a_{2} = 4$, then $a_{3} / b_{3} = 15 / 8$, which is impossible;
(iii) $a_{1} = 2, a_{2} = 5$, then $a_{3} / b_{3} = 3 / 2$, which holds only if $a_{3} = 6$.

The only possible sizes of the box are therefore $(2,3,5)$ and $(2,5,6)$.