Answer: Option D
Let ABC be the equilateral triangle and AD be the altitude on base BC
In an equilateral triangle, the altitude and the median coincide.
So, BC = DC = $$\left( {\frac{{2\sqrt 3 }}{2}} \right)$$ cm = $$\sqrt 3 $$ cm
Let the length of the altitude AD be x cm
Then, in right angled ΔADB,
AB2 = AD2 + BD2
⇒ $${\left( {2\sqrt 3 } \right)^2}$$  = x2 + $${\left( {\sqrt 3 } \right)^2}$$
⇒ x2 = (12 - 3)
⇒ x2 = 9
⇒ x = 3 cm