Answer
$e=\dfrac{2}{\sqrt 8} \approx 0.71$
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 $a=\sqrt 8$ and $b=2$.
Now, $c^2=a^2-b^2 \implies c^2 =(\sqrt 8)^2-(2)^2 \implies c^2= 4$
or, $c=2$
The eccentricity of the ellipse can be found as: $e=\dfrac{c}{a}$
Therefore, $e=\dfrac{2}{\sqrt 8} \approx 0.71$
