Solution:

Let the radius of the circle be $r.$

To find the area of the triangle outside of the circle, we can find the area of the triangle inside the circle and subtract it.

We get that $\angle BOC = 120^{\circ}$ since $\angle ABO$ and $\angle ACO$ are right angles.

This means that the area of sector $OBC$ is $$\dfrac{120}{360} \cdot \pi r^2 = \dfrac{\pi r^2}{3}.$$

Now, we need to find the area of $\triangle BOC.$ Using the formula for the area of a triangle with sine, we get the area to be $$\dfrac{1}{2} \sin (120^{\circ}) \cdot r^2 = \dfrac{r^2\sqrt{3}}{4}.$$

Then the area of the triangle inside the circle is $$\dfrac{\pi r^2}{3} - \dfrac{r^2\sqrt{3}}{4} = \dfrac{r^2(4\pi - 3\sqrt{3})}{12}.$$

The area of $\triangle ABC$ is $$\dfrac{(r\sqrt{3})^2\sqrt{3}}{4} = \dfrac{3r^2\sqrt{3}}{4}.$$

The desired fraction is then $$\begin{align*} 1 - \dfrac{\frac{r^2(4\pi - 3\sqrt{3})}{12}}{\frac{3r^2\sqrt{3}}{4}} &= 1 - \dfrac{4\pi - 3\sqrt{3}}{9\sqrt{3}} \\&= 1 - \dfrac{4\pi\sqrt{3}}{27} + \dfrac{1}{3} \\&= \dfrac{4}{3} - \dfrac{4\pi\sqrt{3}}{27}. \end{align*}$$

Thus, E is the correct answer.