Solution:

We get that the area of the triangle is $$\dfrac{1}{2} \cdot 2\sqrt{3} \cdot 6 = 6\sqrt{3}.$$ The length of the hypotenuse is $$\sqrt{(2\sqrt{3})^2 + 6^2} = \sqrt{48} = 4\sqrt{3}.$$

Dropping the altitude, $h,$ from the vertex to the hypotenuse, we get that $$\dfrac{1}{2} \cdot h \cdot 4\sqrt{3} = 6\sqrt{3}$$$$h = 3.$$

Thus, C is the correct answer.