To solve the problem, we need to analyze the cube assembly, painting, disassembly, and reassembly process.

Step-by-Step Analysis:

1. Initial Cube Construction:

   Pedro assembled a cube using 64 smaller white cubes. Since $64 = 4^3$, the cube is $4 \times 4 \times 4$ in size.

2. Painting the First Cube:

   The first cube is completely painted red. After painting, all external faces of the smaller cubes are red. The only unpainted (white) faces are those inside the cube.

3. Disassembling and Reassembling:

   Pedro disassembles the cube and then reassembles it such that all its faces are white. This implies that he constructs the cube in such a way that all the previously external faces (painted red) are now inside the cube.

4. Painting the Reassembled Cube:

   The reassembled cube is painted red again. Since the entire cube's outer surface is now red, the task is to find out how many smaller cube faces remain white inside.

Calculation:

First Cube (4x4x4):

• Total Number of Small Cubes: $4 \times 4 \times 4 = 64$.
• Cubes on the Surface: The total number of cubes forming the surface can be calculated by excluding the cubes inside:
  • Total number of cubes on the surface: $4 \times 4 \times 6 = 96$ (because there are 6 faces in a cube, and each face has $4 \times 4 = 16$ smaller cubes)
  • However, this counts corner cubes and edge cubes multiple times.
• Cubes inside (not on the surface): These cubes are completely inside and have no exposed faces. This is a $2 \times 2 \times 2 = 8$ cube at the center.

Second Cube Painting:

• When reassembled, to have all external faces white, the red-painted faces must be inside.

• Since the first painting leaves $16$ red cubes on each face and $(4 - 2)^3 = 8$ completely white interior cubes, after the second painting:

• Faces remaining white (completely inside): Only the completely internal $2 \times 2 \times 2$ cube retains its white faces after both paintings.

5. Number of White Faces After Second Painting:

   For the $2 \times 2 \times 2$ cube in the center, there are 8 cubes, each having 6 faces, contributing $8 \times 6 = 48$ total faces.

   After the first painting, these 8 cubes retained all 6 faces white. In the second assembly, these central cubes remain unchanged because their faces are not exposed. Hence, no face from the center cubes remains visible, so:

   Total number of white faces (not visible in the exterior): 192.

Hence, the correct answer is (E) 192.

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If you have any questions or would like more details, feel free to ask!

Here are some related questions to consider:

1. How many cubes have exactly one face painted red in the final assembly?
2. How many cubes have exactly two faces painted red in the final assembly?
3. What is the total surface area of the painted cube after the second painting?
4. How does the painting method affect the count of unpainted cube faces?
5. What would change if the original cube were $5 \times 5 \times 5$ instead of $4 \times 4 \times 4$?
6. How many cubes were painted on all three faces after the first painting?
7. How many internal cubes remain completely unpainted after the second assembly?
8. Can the problem be generalized for a cube of size $n \times n \times n$?

Tip: Consider visualizing the cube's layers to better understand how painting affects each layer of cubes.