1. Identify the ellipse and its major axis

   The ellipse is $\frac{x^2}{a^2}+\frac{y^2}{4}=1,\qquad a>2.$

   Since $a>2$, the semi-major axis is $a$ along the $x$-axis, and the semi-minor axis is $2$ along the $y$-axis.

   So the ends of the major axis are $(a,0) \quad \text{and} \quad (-a,0).$

2. Set up the triangle

   One vertex is at one end of the major axis. Take it to be $A=(a,0).$

   One side of the triangle is parallel to the $y$-axis. Since $A$ is already one vertex, the natural vertical side is the side joining two points of the ellipse with the same $x$-coordinate.

   Let the other two vertices be $B=(x,y),\qquad C=(x,-y),$ where $(x,y)$ lies on the ellipse. Then $BC$ is parallel to the $y$-axis.

3. Area of the triangle

   The base $BC$ has length $BC=2y.$

   The perpendicular distance from $A=(a,0)$ to the vertical line $x=\text{constant}=x$ is $a-x.$

   Hence the area is $\Delta=\frac12\cdot 2y\cdot (a-x)=y(a-x).$

4. Use the ellipse equation

   Since $(x,y)$ lies on the ellipse, $\frac{x^2}{a^2}+\frac{y^2}{4}=1.$

   So $y=2\sqrt{1-\frac{x^2}{a^2}}.$

   Therefore, $\Delta(x)=2(a-x)\sqrt{1-\frac{x^2}{a^2}}.$

5. Simplify using a substitution

   Let $t=\frac{x}{a},\qquad -1\le t\le 1.$

   Then $x=at,$ and $\Delta=2a(1-t)\sqrt{1-t^2}.$

   Now, $\sqrt{1-t^2}=\sqrt{(1-t)(1+t)}.$

   Thus, $\Delta=2a(1-t)\sqrt{(1-t)(1+t)}=2a(1-t)^{3/2}(1+t)^{1/2}.$

6. Maximize the area

   It is easier to maximize $f(t)=(1-t)^3(1+t),$ since $\Delta^2\propto f(t)$.

   Expand: $f(t)=(1-t)^3(1+t).$

   Differentiate using product rule: $f'(t)=(-3)(1-t)^2(1+t)+(1-t)^3.$

   Factor: $f'(t)=(1-t)^2\big[-3(1+t)+(1-t)\big].$

   $f'(t)=(1-t)^2(-3-3t+1-t)=(1-t)^2(-2-4t).$

   So critical points are $t=1 \quad \text{or} \quad t=-\frac12.$

   At $t=1$, area is $0$, so the maximum occurs at $t=-\frac12.$

   Hence $x=-\frac{a}{2}.$

7. Find the maximum area

   When $t=-\frac12$, $1-t=1+\frac12=\frac32,$ $1-t^2=1-\frac14=\frac34.$

   Therefore, $\Delta_{\max}=2a\left(\frac32\right)\left(\frac{\sqrt3}{2}\right)=\frac{3a\sqrt3}{2}.$

   Given that this maximum area is $6\sqrt3$, $\frac{3a\sqrt3}{2}=6\sqrt3.$

   Cancel $\sqrt3$: $\frac{3a}{2}=6 \implies 3a=12 \implies a=4.$

8. Find the eccentricity

   For the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$ with $a>b$, eccentricity is $e=\sqrt{1-\frac{b^2}{a^2}}.$

   Here, $b=2,\qquad a=4.$

   So $e=\sqrt{1-\frac{4}{16}}=\sqrt{1-\frac14}=\sqrt{\frac34}=\frac{\sqrt3}{2}.$

9. Check options

   • A: $\dfrac{\sqrt3}{2}$ ✔
   • B: $\dfrac12$ ✘
   • C: $\dfrac{1}{\sqrt2}$ ✘
   • D: $\dfrac{\sqrt3}{4}$ ✘

Therefore, the correct option is A.