Step 1: Set up the ellipse picture.
Take the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ with $a>b$. The major axis ends are $(\pm a,0)$ and one minor axis end is $B(0,b)$.
Step 2: Use the $120^\circ$ angle.
The major axis subtends $120^\circ$ at $B$. By symmetry the vertical splits it into two $60^\circ$ angles. In the right triangle with legs $b$ (vertical) and $a$ (horizontal), $\tan 60^\circ=\dfrac{a}{b}$.
Step 3: Get a relation.
\[ \sqrt3=\frac{a}{b}\ \Rightarrow\ a=b\sqrt3. \]
Step 4: Use the latus rectum.
The semi latus rectum is $\dfrac{b^2}{a}=\dfrac{4}{\sqrt3}$. Put $a=b\sqrt3$:
\[ \frac{b^2}{b\sqrt3}=\frac{b}{\sqrt3}=\frac{4}{\sqrt3}\ \Rightarrow\ b=4. \]
Step 5: Find $a$.
\[ a=b\sqrt3=4\sqrt3. \]
Step 6: Add both axis lengths.
Major axis $=2a=8\sqrt3$, minor axis $=2b=8$. Sum:
\[ 8\sqrt3+8=8(\sqrt3+1). \]
\[ \boxed{8(\sqrt3+1)} \]