A cube is assembled with $27$ white cubes. The larger cube is then painted black on the outside and disassembled. A blind man reassembles it. What is the probability that the cube is now completely black on the outside? Give an approximation of the size of your answer.
1. **Determine the number of specific cubes and their positions:**
   - There are 27 smaller cubes in the larger cube.
   - These smaller cubes can be categorized as follows:
     - 8 corner cubes
     - 12 edge cubes
     - 6 face-center cubes
     - 1 center cube

2. **Calculate the probability of correct positioning:**
   - The 8 corner cubes must be placed in the 8 corner positions.
   - The 12 edge cubes must be placed in the 12 edge positions.
   - The 6 face-center cubes must be placed in the 6 face-center positions.
   - The 1 center cube must be placed in the center position.
   - The total number of ways to arrange 27 cubes is \(27!\).
   - The number of ways to arrange the cubes correctly is \(8! \cdot 12! \cdot 6! \cdot 1!\).
   - Therefore, the probability of correct positioning is:
     \[
     \frac{8! \cdot 12! \cdot 6! \cdot 1!}{27!}
     \]

3. **Calculate the probability of correct orientation:**
   - Each corner cube has 3 black faces, and it must be oriented such that one of these faces is outward. There are 8 possible orientations for each corner cube, so the probability for each corner cube is \(\frac{1}{8}\).
   - Each edge cube has 2 black faces, and it must be oriented such that one of these faces is outward. There are 12 possible orientations for each edge cube, so the probability for each edge cube is \(\frac{1}{12}\).
   - Each face-center cube has 1 black face, and it must be oriented such that this face is outward. There are 6 possible orientations for each face-center cube, so the probability for each face-center cube is \(\frac{1}{6}\).
   - The center cube can be oriented in any way, so its probability is 1.
   - Therefore, the probability of correct orientation is:
     \[
     \left(\frac{1}{8}\right)^8 \cdot \left(\frac{1}{12}\right)^{12} \cdot \left(\frac{1}{6}\right)^6 \cdot 1
     \]

4. **Combine the probabilities:**
   - The total probability is the product of the probability of correct positioning and the probability of correct orientation:
     \[
     \frac{8! \cdot 12! \cdot 6! \cdot 1!}{27!} \cdot \frac{1}{8^8 \cdot 12^{12} \cdot 6^6}
     \]

5. **Simplify the expression:**
   - Using the factorial and power calculations, we get:
     \[
     \frac{8! \cdot 12! \cdot 6! \cdot 1!}{27!} \cdot \frac{1}{8^8 \cdot 12^{12} \cdot 6^6} = \frac{8! \cdot 12! \cdot 6!}{27! \cdot 8^8 \cdot 12^{12} \cdot 6^6}
     \]

6. **Approximate the probability:**
   - The exact value of the probability is:
     \[
     \frac{8! \cdot 12! \cdot 6!}{27! \cdot 8^8 \cdot 12^{12} \cdot 6^6} \approx 1.83 \times 10^{-37}
     \]

The final answer is \(\boxed{1.83 \times 10^{-37}}\).