A probability problem on limit theorem (!) Each of the 300 workers of a factory takes his lunch in one of the three competing restaurants (equally likely, so with probability $1/3$). How many seats should each restaurants have so that, on average, at most one in 20 customers will remain unseated?
How can I approach this?

It is not clear to me why it is given as an excercise to the limit theorems chapter. 

That this is an exercise on the limit theorems chapter is actually a clue. Let $n=300$ denote the number of workers, $s$ the number of seats in each restaurant and $X$ the number of workers trying to get a seat in restaurant 1. Then $X-s$ customers remain unseated at restaurant 1 if $X\gt s$, and none if $X\leqslant s$ hence the mean number of customers who remain unseated at restaurant 1 is $E[X-s;X\gt s]$. By symmetry, the mean number of customers who remain unseated at any of the three restaurants is $m=3\cdot E[X-s;X\gt s]$ and one asks for $s$ such that $m\leqslant20$.
The distribution of $X$ is binomial $(n,\frac13)$ hence its mean is $\frac13n$ and its variance is $\frac29n$. If $n$ is large enough for the central limit theorem to apply, then 
$$
X=\tfrac13n+\tfrac13\sqrt{2n}\cdot Y,
$$ 
where $Y$ is roughly standard normal. Defining 
$$
s=\tfrac13n+\tfrac13\sqrt{2n}\cdot r,
$$ one gets $m=\sqrt{2n}\cdot E[Y-r;Y\gt r]$. Since $Y$ is approximately standard normal, for every $r$, 
$$
E[Y-r;Y\gt r]\approx\varphi(r)-r(1-\Phi(r)),
$$ 
hence one solves numerically $\varphi(r)-r(1-\Phi(r))=\frac{20}{\sqrt{2n}}=\frac13\sqrt6$. The root is $r^*\approx-0.665$, which yields $s^*=\frac13n+\frac13\sqrt{2n}\cdot r^*\approx100-\frac{20}9\sqrt6\approx94.56$. 
Finally, if the gaussian approximation applies, each restaurant should have at least $95$ seats.
