Solution:

The answer is (D). Let us consider all the non-degenerate triangles whose vertices are also vertices of the hexagon, distinguishing them into 3 types:

Three consecutive vertices. There are $6$ triangles of this type (one for each vertex), and the area of each is equal to $1/6$ of that of the hexagon (see figure).

One vertex every two. There are two triangles of this type, and their area is equal to that of the hexagon minus the area of three small triangles of the first type, that is $1/2$.

Two adjacent vertices and one not. In this case we have $12$ possible triangles, two for each side (once a side is fixed we have two vertices on the opposite side to choose from), and the area of each is equal to $1/3$ (half the area of the hexagon minus

the area of a triangle of the first type).

The sum of the areas of these triangles is therefore
$$6 \cdot \frac{1}{6} + 2 \cdot \frac{1}{2} + 12 \cdot \frac{1}{3} = 6.$$