jmc

First, we note that $55$, $165$, and $260$ all have a common factor of $5$: $$\begin{array}{rl}55& =5\cdot 11\\ 165& =5\cdot 33\\ 260& =5\cdot 52\end{array}$$An integer $n$ satisfies $55n\equiv 165(\mathrm{mod}260)$ if and only if it satisfies $11n\equiv 33(\mathrm{mod}52)$. (Make sure you see why!)

Now it is clear that $n=3$ is a solution. Moreover, since $11$ and $52$ are relatively prime, the solution is unique $(\mathrm{mod}52)$. If you don't already know why this is the case, consider that we are looking for $n$ such that $11n-33=11(n-3)$ is divisible by $52$; this is true if and only if $n-3$ is divisible by $52$.

Hence all solutions are of the form $3+52k$, where $k$ is an integer. One such solution which is easy to compute is $3+52(20)=1043$. The next-largest solution is $1043-52=991$, so the largest three-digit solution is $\boxed{991}$.