The faces of a box with integer edge lengths are painted green. The box is partitioned into unit cubes. Find the dimensions of the box if the number of unit cubes with no green face is one third of the total number of cubes.
1. Let the dimensions of the box be \(a \times b \times c\). The total number of unit cubes in the box is \(abc\).
2. The number of unit cubes with no green face is \((a-2)(b-2)(c-2)\), as these are the cubes that are not on the outermost layer of the box.
3. According to the problem, the number of unit cubes with no green face is one third of the total number of cubes. Therefore, we have the equation:
   \[
   (a-2)(b-2)(c-2) = \frac{abc}{3}
   \]
4. Rearrange the equation to isolate the ratio:
   \[
   3(a-2)(b-2)(c-2) = abc
   \]
   \[
   3 = \frac{abc}{(a-2)(b-2)(c-2)}
   \]
5. To find the possible dimensions, we need to consider the constraints. If all dimensions \(a, b, c \geq 7\), then:
   \[
   \frac{abc}{(a-2)(b-2)(c-2)} \leq \frac{343}{125} \approx 2.74
   \]
   This is less than 3, which means at least one dimension must be less than or equal to 6.
6. Let's assume \(c \leq 6\) and find possible values for \(c\). We will check \(c = 3, 4, 5, 6\).

   - For \(c = 3\):
     \[
     3(a-2)(b-2) = 3ab
     \]
     \[
     (a-2)(b-2) = ab
     \]
     This equation does not hold for any integer values of \(a\) and \(b\).

   - For \(c = 4\):
     \[
     3(a-2)(b-2) = 4ab
     \]
     \[
     3ab - 6a - 6b + 12 = 4ab
     \]
     \[
     ab - 6a - 6b + 12 = 0
     \]
     Solving this equation for integer values, we get possible dimensions:
     \[
     (a, b, c) = (7, 30, 4), (8, 18, 4), (9, 14, 4), (10, 12, 4)
     \]

   - For \(c = 5\):
     \[
     3(a-2)(b-2) = 5ab
     \]
     \[
     3ab - 6a - 6b + 12 = 5ab
     \]
     \[
     2ab - 6a - 6b + 12 = 0
     \]
     Solving this equation for integer values, we get possible dimensions:
     \[
     (a, b, c) = (5, 27, 5), (6, 12, 5), (7, 9, 5)
     \]

   - For \(c = 6\):
     \[
     3(a-2)(b-2) = 6ab
     \]
     \[
     3ab - 6a - 6b + 12 = 6ab
     \]
     \[
     3ab - 6a - 6b + 12 = 0
     \]
     Solving this equation for integer values, we get possible dimensions:
     \[
     (a, b, c) = (6, 8, 6)
     \]

The final answer is \( \boxed{ (7, 30, 4), (8, 18, 4), (9, 14, 4), (10, 12, 4), (5, 27, 5), (6, 12, 5), (7, 9, 5), (6, 8, 6) } \).