In an ellipse, with centre at the origin,

Given that focus is $(0,5 \sqrt{3}) \Rightarrow|b|>|a|$

Let $b>a>0$ and foci is $(0, \pm b e)$

$\because a^{2}=b^{2}-b^{2} e^{2} \Rightarrow b^{2} e^{2}=b^{2}-a^{2}$

$b e=\sqrt{b^{2}-a^{2}} \Rightarrow b^{2}-a^{2}=75$.....(i)

$\because 2 b-2 a=10 \Rightarrow b-a=5 \quad \ldots$ (ii)

From (i) and (ii)

$b+a=15$......(iii)

On solving (ii) and (iii), we get

$\Rightarrow b=10, a=5$

Now, length of latus rectum $=\frac{2 a^{2}}{b}=\frac{50}{10}=5$