Let the maximum area of the triangle that can be inscribed in the ellipse $$\frac{x^2}{a^2} + \frac{y^2}{4} = 1, a > 2$$, having one of its vertices at one end of the major axis of the ellipse and one of its sides parallel to the $$y$$-axis, be $$6\sqrt{3}$$. Then the eccentricity of the ellipse is:

The ellipse is $$\frac{x^2}{a^2} + \frac{y^2}{4} = 1$$ with $$a > 2$$, so the major axis is along the x-axis with semi-major axis $$a$$ and semi-minor axis $$b = 2$$. One vertex of the triangle is at one end of the major axis, say $$(a, 0)$$, and the other two vertices lie on the vertical line $$x = t$$.

From the ellipse equation at $$x = t$$ one gets $$y = \pm 2\sqrt{1 - \frac{t^2}{a^2}}$$. The area of the triangle with vertices $$(a, 0)$$, $$(t, y_0)$$, $$(t, -y_0)$$ is $$A = \frac{1}{2}|a - t|\times 2y_0 = (a - t)\times 2\sqrt{1 - \frac{t^2}{a^2}}$$. Let $$u = \frac{t}{a}$$, so $$A = 2a(1 - u)\sqrt{1 - u^2}$$.

We maximize $$A^2 = 4a^2(1-u)^2(1-u^2) = 4a^2(1-u)^3(1+u)$$. Let $$f(u) = (1-u)^3(1+u)$$; setting $$f'(u) = 0$$ gives

$$f'(u) = -3(1-u)^2(1+u) + (1-u)^3 = (1-u)^2[-3(1+u) + (1-u)] = (1-u)^2(-2 - 4u) = 0$$

Hence $$u = -\frac{1}{2}$$ (since $$u \neq 1$$), meaning $$t = -\frac{a}{2}$$. Computing the maximum area at $$u = -\frac{1}{2}$$ yields

$$f\left(-\frac{1}{2}\right) = \left(\frac{3}{2}\right)^3 \times \frac{1}{2} = \frac{27}{16},\quad A_{\max}^2 = 4a^2 \times \frac{27}{16} = \frac{27a^2}{4},\quad A_{\max} = \frac{3a\sqrt{3}}{2}$$.

Setting $$A_{\max} = 6\sqrt{3}$$ gives

$$\frac{3a\sqrt{3}}{2} = 6\sqrt{3} \implies a = 4$$.

Calculating the eccentricity: $$e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{4}{16}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$$.

Therefore, the eccentricity is Option A: $$\frac{\sqrt{3}}{2}$$.