12. (10 points) A cube with an edge length of 4 contains a sphere with a radius of 1. The sphere can move freely inside the cube, and the cube can be flipped over. The area of the inner surface of the cube that the sphere cannot touch is $\qquad$
【Analysis】First, consider one face alone. Since the radius of the sphere is 1, the outermost 1 unit wide ring of this face is unreachable by the sphere. The sphere can only touch the central 2 unit length square part, with a contact area of 4 square units. The unreachable part is $16-4=12$. This can solve the problem.

【Solution】Solution: First, consider one face alone. Since the radius of the sphere is 1, the outermost 1 unit wide ring of this face is unreachable by the sphere. The sphere can only touch the central 2 unit length square part, with a contact area of 4 square units. The unreachable part is $16-4=12$. For six faces, the total is $12 \times 6=72$ square units. Therefore, the answer is: 72.