#### Answer

$30$ miles

#### Work Step by Step

distance = (rate)$\times$(time) $\Rightarrow$ time = $\displaystyle \frac{\text{distance}}{\text{rate}}$
We are given:$\displaystyle \quad t_{2}=t_{1}+\frac{1}{6}$ (in hours),
where the indices 1 and 2 represent boats 1 and 2.
The time $t_{1}$ is shorter for the faster boat (boat 1)
Let d be the distance (length of the course).
$\displaystyle \frac{d}{18}=\frac{d}{20}+\frac{1}{6}\qquad $.../$\times$LCD= $180$
$10d=9d+30\qquad.../-9d$
$d=30$ miles