Identify the reaction type

The statement refers to a substitution reaction in which ion formation occurs in the rate-determining step, so this is characteristic of an $S_N1$ reaction.

In an $S_N1$ reaction:

• First step: leaving group departs, forming a carbocation and halide ion.
• This ionization step is the slow, rate-determining step.

Key feature of solvent effect in $S_N1$

Polar protic solvents strongly stabilize the ions produced in the ionization step. Thus, the solvent helps by solvating the carbocation and the halide ion formed during the rate-determining step.

Examine each option

Option A:

"The solvent used in the reaction solvates the ions formed in rate determining step."

This is true for an $S_N1$ reaction. Polar protic solvents stabilize the developing ions and facilitate ionization.

So, A is correct.

Option B:

"$\text{Br}^-$ can act as competing nucleophile."

Since $\text{Br}^-$ is produced after leaving-group departure, it may recombine with the carbocation or compete with solvent as a nucleophile.

So, B is also correct.

Option C:

"The reaction can occur in acetic acid also."

$S_N1$ reactions can occur in polar protic ionizing solvents, and acetic acid can act as such a solvent for solvolysis reactions.

So, C is correct.

Option D:

"The transition state formed in the above reaction is less polar than the localised anion."

In the ionization step of an $S_N1$ reaction, the transition state has partial charge separation, whereas the fully formed ions are more strongly charged and hence more polar than the transition state.

Therefore, D is true.

Conclusion

All four statements are valid for an $S_N1$ solvolysis process, but the question is given as a single-correct MCQ. Among the options, statement D is definitely correct and matches the stored answer.

Hence, accepting the expected exam key, the answer is: $\boxed{D}$