Let $$e_1$$ be the eccentricity of the hyperbola $$\frac{x^2}{16} - \frac{y^2}{9} = 1$$ and $$e_2$$ be the eccentricity of the ellipse $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$, $$a > b$$, which passes through the foci of the hyperbola. If $$e_1 e_2 = 1$$, then the length of the chord of the ellipse parallel to the x-axis and passing through $$(0, 2)$$ is :

Hyperbola: $$\frac{x^2}{16} - \frac{y^2}{9} = 1$$. Here $$a_h = 4, b_h = 3$$.

$$e_1 = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$$

Foci of hyperbola: $$(\pm 5, 0)$$.

Given $$e_1 e_2 = 1$$: $$e_2 = \frac{4}{5}$$.

Ellipse passes through $$(\pm 5, 0)$$:

$$\frac{25}{a^2} = 1 \Rightarrow a^2 = 25, a = 5$$.

$$e_2 = \sqrt{1 - \frac{b^2}{a^2}} = \frac{4}{5}$$

$$\frac{b^2}{25} = 1 - \frac{16}{25} = \frac{9}{25}$$

$$b^2 = 9$$

Ellipse: $$\frac{x^2}{25} + \frac{y^2}{9} = 1$$.

Chord at $$y = 2$$: $$\frac{x^2}{25} + \frac{4}{9} = 1 \Rightarrow \frac{x^2}{25} = \frac{5}{9} \Rightarrow x^2 = \frac{125}{9}$$

$$x = \pm\frac{5\sqrt{5}}{3}$$

Length = $$\frac{10\sqrt{5}}{3}$$.

The answer corresponds to Option (3).