12. (10 points) In a sphere with a radius of 10 cm, there is a cube with integer (cm) edge lengths. What is the maximum edge length of the cube?

 untranslated part:
设半径为 10 厘米的球中有一个棱长为整数 (厘米) 的正方体, 则该正方体的棱长最大等于多少？

translated part:
In a sphere with a radius of 10 cm, there is a cube with integer (cm) edge lengths. What is the maximum edge length of the cube?
【Analysis】Draw the figure, see the figure below, the vertices of the inscribed cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ in the sphere are on the sphere's surface, its (body) diagonal $A C_{1}$ is the diameter of the sphere, i.e., $A C_{1}=2 \times 10=20$ (cm). Then, by the symmetry of the figure and the Pythagorean theorem, solve the problem.

【Solution】Solution:
The vertices of the inscribed cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ in the sphere are on the sphere's surface, its (body) diagonal $A C_{1}$ is the diameter of the sphere, i.e., $A C_{1}=2 \times 10=20$ (cm).

By the symmetry of the figure, we know $\angle A A_{1} C_{1}=90^{\circ}, \angle A_{1} B_{1} C_{1}=90^{\circ}$.
Let the edge length of the cube be $a$, i.e., $A A_{1}=A_{1} B_{1}=B_{1} C_{1}=a$. Using the Pythagorean theorem twice, we get:
$$
A_{1} C_{1}^{2}=2 a^{2}, A C_{1}^{2}=A A_{1}^{2}+A 1 C_{1}^{2}+3 a^{2} \text {, }
$$

Then $3 a^{2}=20^{2}=400, a^{2}=\frac{400}{3}=133 \frac{1}{3}$.
Clearly, as long as the edge length $a$ of a cube is an integer, and satisfies $a^{2} \leqslant 133$, then this cube can definitely fit into the sphere, because $11^{2}=121<133<144=12^{2}$. Therefore, the maximum edge length of the cube with an integer edge length is 11 cm.

Answer: The maximum edge length of the cube is 11 cm.