1. Find the eccentricity of the hyperbola

Given hyperbola: $\frac{x^2}{16}-\frac{y^2}{9}=1$

Here, $a^2=16,\quad b^2=9$

For the hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$, we have $c^2=a^2+b^2=16+9=25 \implies c=5$

So its eccentricity is $e_1=\frac{c}{a}=\frac{5}{4}$

Its foci are $(\pm 5,0)$

2. Use the condition $e_1e_2=1$ to find $e_2$

Given $e_1e_2=1$ So, $e_2=\frac{1}{e_1}=\frac{4}{5}$

3. Form the ellipse using its eccentricity and the fact that it passes through the foci of the hyperbola

Ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\quad a>b$

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

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

Now the ellipse passes through $(5,0)$ (and also $(-5,0)$), since these are the foci of the hyperbola.

Substitute $(5,0)$ into ellipse: $\frac{25}{a^2}=1 \implies a^2=25$

Hence, $b^2=\frac{9}{25}\cdot 25=9$

So the ellipse is $\frac{x^2}{25}+\frac{y^2}{9}=1$

4. Find the chord parallel to the $x$-axis through $(0,2)$

A chord parallel to the $x$-axis through $(0,2)$ is the line $y=2$

Substitute into the ellipse: $\frac{x^2}{25}+\frac{4}{9}=1$ $\frac{x^2}{25}=1-\frac{4}{9}=\frac{5}{9}$ $x^2=25\cdot \frac{5}{9}=\frac{125}{9}$ $x=\pm \frac{5\sqrt{5}}{3}$

So the chord length is $\left(\frac{5\sqrt{5}}{3}-\left(-\frac{5\sqrt{5}}{3}\right)\right)=\frac{10\sqrt{5}}{3}$

5. Compare with options

Thus the required length is $\boxed{\frac{10\sqrt{5}}{3}}$

So the correct option is D.