$$ \text{The area of the region, inside the circle }(x-2\sqrt{3})^{2}+y^{2}=12 \text{ and outside the parabola } y^{2}=2\sqrt{3}x \text{ is :} $$

The circle $$(x-2\sqrt{3})^{2}+y^{2}=12$$ has centre $$\left(2\sqrt{3},\,0\right)$$ and radius $$r=\sqrt{12}=2\sqrt{3}$$.

The parabola $$y^{2}=2\sqrt{3}\,x$$ opens towards the positive $$x$$-axis and is symmetric about the $$x$$-axis.

Points common to both curves are obtained by substituting $$y^{2}=2\sqrt{3}x$$ in the circle:

$$(x-2\sqrt{3})^{2}+2\sqrt{3}\,x=12 \; \Longrightarrow \; x^{2}-2\sqrt{3}\,x=0 \; \Longrightarrow \; x\bigl(x-2\sqrt{3}\bigr)=0.$$

Hence $$x=0$$ gives the origin $$(0,0)$$, and $$x=2\sqrt{3}$$ gives the points $$(2\sqrt{3},\; \pm 2\sqrt{3})$$. Thus the two curves intersect at the three points $$\bigl(0,0\bigr)$$ and $$\bigl(2\sqrt{3},\pm 2\sqrt{3}\bigr).$$

Because the figure is symmetric about the $$x$$-axis, we calculate the area above the axis and double it.

For a fixed $$y$$ between $$0$$ and $$2\sqrt{3}$$ the left boundary of the circle is obtained from $$(x-2\sqrt{3})^{2}+y^{2}=12 \quad\Rightarrow\quad x=2\sqrt{3}-\sqrt{\,12-y^{2}\,}.$$ The parabola gives $$x=\dfrac{y^{2}}{2\sqrt{3}}.$$ Inside the circle but outside the parabola means the $$x$$-coordinate runs from the circle to the parabola. Hence the required area $$A$$ is

$$A=2\int_{0}^{2\sqrt{3}}\left[\frac{y^{2}}{2\sqrt{3}}-\Bigl(2\sqrt{3}-\sqrt{12-y^{2}}\Bigr)\right]\,dy.$$

Split the integral:

$$A=2\Biggl[\;\underbrace{\int_{0}^{2\sqrt{3}}\sqrt{12-y^{2}}\,dy}_{I_{1}}\;+\;\underbrace{\int_{0}^{2\sqrt{3}}\left(\frac{y^{2}}{2\sqrt{3}}-2\sqrt{3}\right)dy}_{I_{2}}\;\Biggr].$$

Case 1: Evaluate $$I_{1}$$. The integrand represents the upper half of a circle of radius $$2\sqrt{3}$$. Area of a semicircle $$=\frac{1}{2}\pi r^{2}=\frac{1}{2}\pi(2\sqrt{3})^{2}=6\pi$$. Taking the upper quarter (from $$y=0$$ to $$y=2\sqrt{3}$$) gives $$I_{1}=3\pi.$$

Case 2: Evaluate $$I_{2}$$.

$$\int_{0}^{2\sqrt{3}}\frac{y^{2}}{2\sqrt{3}}\,dy=\frac{1}{2\sqrt{3}}\cdot\frac{(2\sqrt{3})^{3}}{3}=4,$$ $$\int_{0}^{2\sqrt{3}}2\sqrt{3}\,dy=2\sqrt{3}\cdot2\sqrt{3}=12.$$ Therefore $$I_{2}=4-12=-8.$$

Combine the two parts:

$$A=2\bigl(I_{1}+I_{2}\bigr)=2\bigl(3\pi-8\bigr)=6\pi-16.$$

Hence the required area is $$6\pi-16$$, which matches Option B.