The area (in sq. units) of an equilateral triangle inscribed in the parabola $$y^2 = 8x$$, with one of its vertices on the vertex of this parabola is:

We consider the parabola $$y^2 = 8x$$ whose vertex is at the origin. Let us place one vertex of the desired equilateral triangle at this point, calling it $$A(0,0).$$ Because the curve is symmetric about the $$x$$-axis, it is natural to look for the other two vertices as a pair of points that are mirror images with respect to this axis. Hence we write

$$B(x_0,\;y_0)\quad\text{and}\quad C(x_0,\;-y_0)$$

with $$y_0 > 0$$ and $$B,\,C$$ lying on the parabola. So we have

$$y_0^2 = 8x_0. \quad -(1)$$

Now an equilateral triangle requires all three sides to be equal, so we equate the squared lengths:

Length $$AB^2 = (x_0-0)^2 + (y_0-0)^2 = x_0^2 + y_0^2,$$

Length $$BC^2 = (x_0 - x_0)^2 + (y_0 + y_0)^2 = 0^2 + (2y_0)^2 = 4y_0^2.$$

Setting $$AB = BC$$ gives

$$x_0^2 + y_0^2 = 4y_0^2.$$

Simplifying,

$$x_0^2 = 3y_0^2. \quad -(2)$$

From (2) we obtain $$x_0 = \sqrt{3}\,y_0.$$ Substituting this value of $$x_0$$ into the parabola condition (1) yields

$$y_0^2 = 8(\sqrt{3}\,y_0).$$

Dividing by $$y_0$$ (which is positive) we get

$$y_0 = 8\sqrt{3}.$$

Putting this back into $$x_0 = \sqrt{3}\,y_0$$ gives

$$x_0 = \sqrt{3}\,(8\sqrt{3}) = 8\cdot3 = 24.$$

Hence the coordinates of the three vertices are

$$A(0,0),\; B(24,\,8\sqrt{3}),\; C(24,\,-8\sqrt{3}).$$

The side length can now be found. Using $$AB$$,

$$AB^2 = 24^2 + (8\sqrt{3})^2 = 576 + 64\cdot3 = 576 + 192 = 768,$$

so

$$AB = \sqrt{768} = \sqrt{256\cdot3} = 16\sqrt{3}.$$

The standard formula for the area of an equilateral triangle with side $$s$$ is

$$\text{Area} = \frac{\sqrt{3}}{4}\,s^2.$$

Substituting $$s = 16\sqrt{3}$$ we have

$$\text{Area} = \frac{\sqrt{3}}{4}\,(16\sqrt{3})^2 = \frac{\sqrt{3}}{4}\,(256\cdot3) = \frac{768\sqrt{3}}{4} = 192\sqrt{3}\ \text{square units}.$$

Hence, the correct answer is Option C.