Grogg takes an a × b × c rectangular block (where a, b, c are positive integers), paints
the outside of it purple, and cuts it into abc small 1 × 1 × 1 cubes. He then places all the
small cubes into a bag, and Winnie reaches in and randomly picks one of the small cubes.
If the probability that Winnie picks a totally unpainted cube is 20%, determine all possible
values of the number of cubes in the bag.
1. **Assume \( a \le b \le c \):** 
   Since \( a, b, c \ge 3 \) (if any of them were less than or equal to 2, there would be no unpainted cubes), we can proceed with the assumption \( a \le b \le c \).

2. **Probability of picking an unpainted cube:**
   A cube is unpainted if and only if it is part of the interior \((a-2) \times (b-2) \times (c-2)\) cube inside the larger cube. Thus, the probability that a cube is unpainted is:
   \[
   \frac{(a-2)(b-2)(c-2)}{abc} = \frac{1}{5}
   \]

3. **Claim: \( a \le 4 \):**
   Assume otherwise, i.e., \( a, b, c \ge 5 \). Then:
   \[
   \frac{(a-2)(b-2)(c-2)}{abc} = \left(1 - \frac{2}{a}\right)\left(1 - \frac{2}{b}\right)\left(1 - \frac{2}{c}\right)
   \]
   Noting that \( \frac{2}{a} \le \frac{2}{5} \), we see that \( 1 - \frac{2}{a} \ge \frac{3}{5} \), and similarly for \( b \) and \( c \). Therefore:
   \[
   \frac{(a-2)(b-2)(c-2)}{abc} \ge \left(\frac{3}{5}\right)^3 = \frac{27}{125} > \frac{1}{5}
   \]
   This is a contradiction. Hence, \( a \le 4 \).

4. **Case 1: \( a = 3 \):**
   \[
   \frac{(3-2)(b-2)(c-2)}{3bc} = \frac{(b-2)(c-2)}{3bc} = \frac{1}{5}
   \]
   Cross-multiplying, we get:
   \[
   5(b-2)(c-2) = 3bc
   \]
   Expanding and simplifying:
   \[
   5bc - 10b - 10c + 20 = 3bc
   \]
   \[
   2bc - 10b - 10c + 20 = 0
   \]
   Adding 15 to both sides and factoring:
   \[
   (b-5)(c-5) = 15
   \]
   Considering \( b - 5 \le c - 5 \):
   - If \( b - 5 = 1 \) and \( c - 5 = 15 \), then \( b = 6 \) and \( c = 20 \). Total cubes: \( 3 \cdot 6 \cdot 20 = 360 \).
   - If \( b - 5 = 3 \) and \( c - 5 = 5 \), then \( b = 8 \) and \( c = 10 \). Total cubes: \( 3 \cdot 8 \cdot 10 = 240 \).
   - Other cases lead to negative or zero values for \( b \) or \( c \), which are not valid.

5. **Case 2: \( a = 4 \):**
   \[
   \frac{(4-2)(b-2)(c-2)}{4bc} = \frac{(b-2)(c-2)}{2bc} = \frac{1}{5}
   \]
   Cross-multiplying, we get:
   \[
   5(b-2)(c-2) = 2bc
   \]
   Expanding and simplifying:
   \[
   5bc - 10b - 10c + 20 = 2bc
   \]
   \[
   3bc - 10b - 10c + 20 = 0
   \]
   Adding 40 to both sides and factoring:
   \[
   (3b-10)(3c-10) = 40
   \]
   Considering \( 3b - 10 \le 3c - 10 \):
   - If \( 3b - 10 = 2 \) and \( 3c - 10 = 20 \), then \( b = 4 \) and \( c = 10 \). Total cubes: \( 4 \cdot 4 \cdot 10 = 160 \).
   - If \( 3b - 10 = 5 \) and \( 3c - 10 = 8 \), then \( b = 5 \) and \( c = 6 \). Total cubes: \( 4 \cdot 5 \cdot 6 = 120 \).

Therefore, the only possible values of the number of cubes in the bag are \( \boxed{120, 160, 240, 360} \).