1. Identify the reagent generated in situ

A mixture of $KBr$ and $KBrO_3$ in solution generates bromine under acidic conditions:

$BrO_3^- + 5Br^- + 6H^+ \rightarrow 3Br_2 + 3H_2O$

So effectively, phenol is being treated with bromine water / brominating conditions.

2. Reactivity of phenol toward electrophilic substitution

The $-OH$ group in phenol is strongly activating due to resonance donation of lone pair electrons into the benzene ring. Therefore:

• the ring becomes highly reactive toward electrophilic substitution,
• substitution occurs mainly at the ortho and para positions.

Because phenol is very strongly activating, bromination does not usually stop at mono-bromination under such conditions.

3. Effect of bromine on phenol

When phenol reacts with bromine water, it undergoes rapid multiple substitution to give:

$\text{Phenol} \xrightarrow{Br_2} \text{2,4,6-tribromophenol}$

The three activated positions ($2,4,6$) all get brominated.

4. Evaluate the options

• A: 3-Bromophenol — incorrect, meta substitution is not favored.
• B: 4-Bromophenol — incorrect as major product under these conditions because bromination continues further.
• C: 2,4,6-Tribromophenol — correct.
• D: 2-Bromophenol — incorrect as major product for the same reason.

5. Final answer

The major product is:

$\boxed{\text{2,4,6-Tribromophenol}}$

So, the correct option is C.