Answer
This reaction will proceed via $S_N1$.
Work Step by Step
1. Substrate
- The carbon connected to the leaving group is connected to 3 other carbons.
This favors $S_N1$ reactions.
2. Nucleophile
- $Br^-$ is a strong* nucleophile, which favors $S_N2$ reactions.
*(You can consult the table on page 216)
3. Leaving group
- In this case, the "$OH^-$" is protonated by the $HBr$, and it is converted into a good leaving group (page 218), which favors $S_N2$
4. Solvent
- Not mentioned: not a factor.
- In this case, since the substrate has higher priority, this reaction will occur via $S_N1$
