We are given an ellipse

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

and a hyperbola

$$ \frac{x^2}{A^2}-\frac{y^2}{B^2}=1. $$

It is stated that “the distance between the foci of $E$ and the foci of $H$ is $2\sqrt{3}$.” A natural interpretation is that the foci of the ellipse are separated by

$$ 2\sqrt{a^2-b^2}, $$

and those of the hyperbola by

$$ 2\sqrt{A^2+B^2}, $$

and each distance is equal to $2\sqrt{3}$. That is, we have

$ 2\sqrt{a^2-b^2}=2\sqrt{3}\quad \Longrightarrow\quad a^2-b^2=3, $

and

$ 2\sqrt{A^2+B^2}=2\sqrt{3}\quad \Longrightarrow\quad A^2+B^2=3. $

We are also given that

$$ a-A=2, $$

and that the ratio of the eccentricities is

$$ \frac{e_E}{e_H}=\frac{1}{3}, $$

where the eccentricity of the ellipse is

$$ e_E=\frac{\sqrt{a^2-b^2}}{a}=\frac{\sqrt{3}}{a}, $$

and the eccentricity of the hyperbola is

$$ e_H=\frac{\sqrt{A^2+B^2}}{A}=\frac{\sqrt{3}}{A}. $$

Thus the ratio becomes

$$ \frac{e_E}{e_H}=\frac{\sqrt{3}/a}{\sqrt{3}/A}=\frac{A}{a}=\frac{1}{3}. $$

This implies

$$ a=3A. $$

Now, using the condition $a-A=2$ together with $a=3A$, we get

$$ 3A-A=2 \quad\Longrightarrow\quad 2A=2 \quad\Longrightarrow\quad A=1. $$

Thus,

$$ a=3. $$

Next, for the ellipse we have

$$ a^2-b^2=3 \quad \Longrightarrow\quad 9-b^2=3 \quad\Longrightarrow\quad b^2=6. $$

For the hyperbola,

$$ A^2+B^2=3 \quad \Longrightarrow\quad 1+B^2=3 \quad\Longrightarrow\quad B^2=2. $$

The length of the latus rectum is given by the following formulas:

For the ellipse:

$$ L_E=\frac{2b^2}{a}, $$

For the hyperbola:

$$ L_H=\frac{2B^2}{A}. $$

Substitute the computed values:

For the ellipse:

$$ L_E=\frac{2\times6}{3}=\frac{12}{3}=4, $$

For the hyperbola:

$$ L_H=\frac{2\times2}{1}=4. $$

The sum of the lengths of the latus rectums is then

$$ L_E+L_H=4+4=8. $$

Thus, the answer is

$$ 8. $$