ahh i do not really remember how to work these, but i will do what i can

we are asked to find the coefficent of $ x^3 $ in $ \left(x+2\sqrt{3}\right)^5 $

in these types of binomials, your exponent power value is going to be the $n$ and the $s$, or $i$, or $a$, i do not know what you know them as is going to be the coefficient thinge -- the one we are looking for is $i=3$

so we know that $n=5$ and $i=3$

plug it all into $ \frac{n!}{i!\left(n-i\right)!}x^{\left(n-i\right)}\left(2\sqrt{3}\right)^i $

as following, we work it out:

$\frac{5!}{3!\left(5-3\right)!}x^2\left(2\sqrt{3}\right)^3 $

$\frac{120}{6\cdot \:2}x^2\left(2\sqrt{3}\right)^3 $

$ 10\left(2\sqrt{3}\right)^3x^2 $

$ 10\cdot \: 8\cdot \: 3^{\frac{1}{2}\cdot \:3} x^2 $

$ 10\cdot \: 8\cdot \: 3^{\frac{3}{2}} x^2 $

$ 10\cdot \: 8\cdot \: 3^{\frac{3}{2}} x^2 $

$ 10\cdot \:8\cdot \:3\sqrt{3}\cdot \: x^2 $

$ \boxed{240\cdot \sqrt{3}\cdot \: x^2 } $

if you want to get into decimals it would be:

$ 240\cdot 1.73\cdot \: x^2 $

$\boxed{ 415.2\cdot \:x^2 } $

:D