1. Let the ellipse be in standard form

   Assume the ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \qquad a>b$

   For this ellipse:

   • Distance between foci $=2c$
   • Eccentricity $e=\frac{c}{a}$
   • Directrices are $x=\pm \frac{a}{e}$
   • Hence distance between directrices is $2\cdot \frac{a}{e}=\frac{2a}{e}$

2. Use the given distance between foci

   Given distance between foci is $6$, so $2c=6 \implies c=3$

3. Use the given distance between directrices

   Given distance between directrices is $12$, so $\frac{2a}{e}=12 \implies \frac{a}{e}=6$

   Since $e=\frac{c}{a}$, $\frac{a}{e}=\frac{a}{c/a}=\frac{a^2}{c}$

   Therefore, $\frac{a^2}{c}=6$

   Substituting $c=3$: $\frac{a^2}{3}=6 \implies a^2=18$

4. Find $b^2$

   For an ellipse, $c^2=a^2-b^2$ so $b^2=a^2-c^2=18-9=9$

5. Find the length of latus rectum

   Length of latus rectum of the ellipse is $\frac{2b^2}{a}$

   Substitute $b^2=9$ and $a=\sqrt{18}=3\sqrt{2}$: $\frac{2\cdot 9}{3\sqrt{2}}=\frac{18}{3\sqrt{2}}=\frac{6}{\sqrt{2}}=3\sqrt{2}$

6. Compare with options

   $3\sqrt{2}$ matches Option B.

Therefore, the length of the latus rectum is $3\sqrt{2}$.