smc

Let the boat travel over still water at rate $r$. Then, it travels downstream at rate $r+3$ and upstream at rate $r-3$. The distance travelled downstream can be expressed as $(r+3)t$, where $t$ is the time taken to travel downstream. Because the total time taken is $1$ hour, the distance travelled upstream can be represented as $(r-3)(1-t)$. Because both distances are the same ($4$ miles), we can equate the two expressions to solve for $t$ in terms of $r$: $$\begin{array}{rl}(r+3)t& =(r-3)(1-t)\\ rt+3t& =r-rt-3+3t\\ 2rt& =r-3\\ t& =\frac{r-3}{2r}\end{array}$$ Recalling that the distance travelled downstream, $(r+3)t$, is $4$ miles, we can substitute the above expression for $t$ to solve for $r$: $$\begin{array}{rl}(r+3)t& =4\\ (r+3)\frac{r-3}{2r}& =4\\ r^2-9& =8r\\ r^2-8r-9& =0\\ (r-9)(r+1)& =0\end{array}$$ Because $r>0$, $r=9$. Thus, the ratio of the downstream rate $r+3$ to the upstream rate $r-3$ is $\frac{9+3}{9-3}=\frac{12}{6}=\boxed{2:1}$.