The structure of the compound that gives a tribromo derivative on treatment with bromine water is

In aqueous medium the reaction of molecular bromine with an aromatic ring is an electrophilic aromatic substitution. Only those rings that are very strongly activated can force bromine to substitute at every ortho and para position and thus give the white precipitate of a tribromo derivative.

The -OH group of phenol donates electron density to the ring by resonance: $$O\!H \;\longrightarrow\; \overset{\scriptsize\!-\!}{O}\,\!\!-\!\!\overset{+}{\!H}$$. This $$+M$$ (mesomeric) effect is so strong that the ortho (2-, 6-) and para (4-) positions become much more electron-rich than bromine itself. Consequently, even the mild reagent “bromine water” substitutes bromine at all three of these positions in one step:

$$\text{C}_6\text{H}_5\text{OH}\;+\;3\,Br_2\;(aq)\;\longrightarrow\;2,4,6\text{-tribromophenol (ppt)}\;+\;3\,HBr$$

None of the other listed compounds can achieve this: their rings are either less activated (e.g. anisole, benzyl alcohol) or even deactivated; they stop after mono- or di-bromination, or substitute outside the ring.

Therefore the only structure that gives a tribromo derivative with bromine water is phenol.

Option A which is: phenol ($$C_6H_5OH$$)