To solve, we need to know that a triangle inscribed in a circle where one side is the diameter will always be a right triangle. This means that ABC is a 30-60-90 triangle, which means its sides must have a ratio of x : x√3 : 2x, where x is BC, x√3 is AB, and 2x is AC. Since AC is the diameter of the circle, the radius of the circle will equal x. We can then use A=πr^2 to solve for the area of the circle.

We know that the area of the triangle is 6 and can be found with A=(1/2)bh, where b and h are BC and AB. Plugging in our area and the values for the sides of the triangle gives $$6=\frac{1}{2}\left(BC\right)\left(AB\right)$$ $$6=\frac{1}{2}\left(x\right)\left(x\sqrt{3}\right)$$ $$6=\frac{\sqrt{3}}{2}x^2$$ $$\frac{12}{\sqrt{3}}=x^2$$ $$4\sqrt{3}=x^2$$

Rather than solving for x, we can now plug 4√3 directly into our circle area equation, since x^2=r^2:
$$A=\pi r^2$$ $$A=\pi\left(4\sqrt{3}\right)$$
which is answer choice E.