The length of the semi-latus rectum of an ellipse is one third of its major axis, its eccentricity would be

Let eq. of ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\text{, }$

length of semi-latus rectum 

$=\frac{b^2}{a}=\frac{a^2\left(1-e^2\right)}{a}=a\left(1-e^2\right)$

Given $a\left(1-e^2\right)=\frac{1}{3}\left(2a\right)$

$\Rightarrow 1-e^2=\frac{2}{3}\Rightarrow e^2=1-\frac{2}{3}=\frac{1}{3}\Rightarrow e=\frac{1}{\sqrt{3}}$