Let the ellipse $$E_{1}:\f\frac{x^{2}}{a^{2}}+\f\frac{y^{2}}{b^{2}}=1,a \gt b$$ and $$E_{2}:\f\frac{x^{2}}{A^{2}}+\f\frac{y^{2}}{B^{2}}=1,A \lt B$$ have same eccentricity $$\f\frac{1}{\sqrt{3}}$$. Let the product of their lengths of latus rectums be $$\f\frac{32}{\sqrt{3}}$$, 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 $$e = \f\frac{1}{\sqrt{3}}$$, product of latus rectum lengths $$= \f\frac{32}{\sqrt{3}}$$, and distance between foci of $$E_1$$ is 4.

For the ellipse $$E_1$$ given by $$\f\frac{x^2}{a^2} + \f\frac{y^2}{b^2} = 1$$ with $$a \gt b$$ and eccentricity $$e = \f\frac{1}{\sqrt{3}}$$, 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\l\left(1 - \f\frac{1}{3}\r\right) = 12 \t\times \f\frac{2}{3} = 8

$$

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

$$

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

$$

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

$$

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

$$

Its latus rectum is

$$

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

$$

Since the product of the latus recta is

$$

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

$$

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

$$

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

$$

The equations of the ellipses become

$$

E_1: \f\frac{x^2}{12} + \f\frac{y^2}{8} = 1,

$$

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

$$

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

$$

y^2 \t\times \l\left(\f\frac{-5}{36}\r\right) = -1 \implies y^2 = \f\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}$$.