Let the ellipse $$E_{1}:\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,a \gt b$$ and $$E_{2}:\frac{x^{2}}{A^{2}}+\frac{y^{2}}{B^{2}}=1,A \lt B$$ have same eccentricity $$\frac{1}{\sqrt{3}}$$. Let the product of their lengths of latus rectums be $$\frac{32}{\sqrt{3}}$$, and the distance between the foci of $$E_{1}$$ be 4. If $$E_{1}$$ and $$E_{2}$$ meet at $$A,B,C$$ and $$D,$$ then the area of the quadrilateral $$ABCD$$ equals:

We have two ellipses with the same eccentricity $$e = \frac{1}{\sqrt{3}}$$, product of latus rectum lengths $$= \frac{32}{\sqrt{3}}$$, and distance between foci of $$E_1$$ is 4.

For the ellipse $$E_1$$ given by $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ with $$a \gt b$$ and eccentricity $$e = \frac{1}{\sqrt{3}}$$, the distance between the foci equals $$2ae = 4$$, so $$ae = 2$$, yielding $$a = 2\sqrt{3}$$. Then

$$b^2 = a^2(1 - e^2) = 12\left(1 - \frac{1}{3}\right) = 12 \times \frac{2}{3} = 8$$

and hence $$b = 2\sqrt{2}$$. The latus rectum of $$E_1$$ is

$$\ell_1 = \frac{2b^2}{a} = \frac{16}{2\sqrt{3}} = \frac{8}{\sqrt{3}}.$$

For the ellipse $$E_2$$ defined by $$\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$$ with $$A \lt B$$ (so the major axis is along the y-axis) and the same eccentricity $$e = \frac{1}{\sqrt{3}}$$, we have

$$A^2 = B^2(1 - e^2) = \frac{2B^2}{3}.$$

Its latus rectum is

$$\ell_2 = \frac{2A^2}{B} = \frac{2 \cdot \frac{2B^2}{3}}{B} = \frac{4B}{3}.$$

Since the product of the latus recta is

$$\ell_1 \cdot \ell_2 = \frac{8}{\sqrt{3}} \cdot \frac{4B}{3} = \frac{32B}{3\sqrt{3}} = \frac{32}{\sqrt{3}},$$

it follows that $$B = 3$$ and hence

$$A^2 = \frac{2 \times 9}{3} = 6.$$

The equations of the ellipses become

$$E_1: \frac{x^2}{12} + \frac{y^2}{8} = 1,\qquad E_2: \frac{x^2}{6} + \frac{y^2}{9} = 1.$$

From the first,

$$x^2 = 12\left(1 - \frac{y^2}{8}\right) = 12 - \frac{3y^2}{2},$$

which substituted into the second gives

$$\frac{12 - \frac{3y^2}{2}}{6} + \frac{y^2}{9} = 1$$

$$2 - \frac{y^2}{4} + \frac{y^2}{9} = 1$$

$$y^2\left(\frac{1}{9} - \frac{1}{4}\right) = -1$$

$$y^2 \times \left(\frac{-5}{36}\right) = -1 \implies y^2 = \frac{36}{5},$$

$$x^2 = 12 - \frac{3}{2} \times \frac{36}{5} = 12 - \frac{54}{5} = \frac{6}{5}.$$

Thus the four intersection points are

$$\left(\pm\sqrt{\frac{6}{5}},\; \pm\sqrt{\frac{36}{5}}\right),$$

forming a rectangle by symmetry. Its side lengths are $$2\sqrt{\frac{6}{5}}$$ and $$2\sqrt{\frac{36}{5}}$$, so the area is

$$2\sqrt{\frac{6}{5}} \times 2\sqrt{\frac{36}{5}} = 4\sqrt{\frac{216}{25}} = 4 \times \frac{6\sqrt{6}}{5} = \frac{24\sqrt{6}}{5}.$$

The correct answer is Option 4: $$\frac{24\sqrt{6}}{5}$$.