92 A cube, with edge length $n(n>2, n \in \mathbf{N})$, is entirely painted black. It is then divided into $n^{3}$ smaller cubes, each with an edge length of 1. If the number of smaller cubes that have exactly one face painted black is equal to the number of smaller cubes that have no faces painted black, then $n$ equals
A. 5
B. 6
C. 7
D. 8
92 D. The number of cubes with exactly one face painted black is $6(n-2)^{2}$. The number of cubes with no faces painted black is $(n-$ $2)^{3}$. From
$$
6(n-2)^{2}=(n-2)^{3}
$$

we get $n=8$.