2. D.

As shown in Figure 4, connect $A D$, and let $A D=2 h$. Then the regular hexagon is composed of six equilateral triangles with side length $h$. It is easy to know that the area of the equilateral $\triangle C O B$ with side length $h$ is $\frac{\sqrt{3}}{4} h^{2}$.
Therefore, $S_{\text {regular hexagon }}=6 \times \frac{\sqrt{3}}{4} h^{2}=\frac{3 \sqrt{3}}{2} h^{2}=135$
$$
\Rightarrow h^{2}=30 \sqrt{3} \text {. }
$$

Let the side length of the rhombus $A M=a$.
Then $h=\frac{\sqrt{3}}{2} a \Rightarrow a^{2}=\frac{4}{3} h^{2}$.
Thus, the area of the rhombus $A M D N$ is
$$
2 \times \frac{\sqrt{3}}{4} a^{2}=\frac{\sqrt{3}}{2} \times \frac{4}{3} h^{2}=\frac{2}{\sqrt{3}} \times 30 \sqrt{3}
$$
$=60$ (square centimeters).
Therefore, the area of the remaining part after cutting out this rhombus is 75 square centimeters.
2. As shown in Figure 1, in a regular hexagon $A B C D E F$ with an area of 135 square centimeters, there is a rhombus $A M D N$ composed of two equilateral triangles. Then the area of the remaining part after cutting out this rhombus is ( ) square centimeters.
(A) 60
(B) 65
(C) 70
(D) 75