Answer
$e=\dfrac{1}{2}$
Work Step by Step
The standard form of an ellipse can be written as: $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1....(1); $and has foci $( \pm c, 0)$ where ; $c^2=a^2-b^2$.
On comparison, we have $a=2$ and $b=\sqrt 3$.
Now, $c^2=a^2-b^2 \implies c^2 =9^2-(\sqrt 3)^2 \implies c^2= 1$
or, $c=1$
The eccentricity can be found as: $e=\dfrac{c}{a}$
Therefore, $e=\dfrac{1}{2}$
