For reaction
The correct order of set of reagents for the above conversion is :

The starting compound is aniline ($$C_6H_5NH_2$$) and the target is a mono-brominated aniline. Because the -$$NH_2$$ group is a very strong ortho/para-directing activator, direct treatment of aniline with $$Br_2$$ would give mainly 2,4,6-tribromo­aniline. Hence we must first protect the amino group, then carry out controlled bromination, and finally remove the protecting group.

Step 1 : Protonation
Concentrated $$H_2SO_4$$ converts aniline into its anilinium hydrogen-sulphate salt $$C_6H_5NH_3^+HSO_4^-$$. This salt is less reactive toward electrophilic substitution and also dissolves readily so that acetylation takes place smoothly in the next step.

Step 2 : Acetylation (protection of $$NH_2$$)
With acetic anhydride ($$Ac_2O$$) the anilinium ion is converted to acetanilide: $$C_6H_5NHCOCH_3$$. The -$$NHCOCH_3$$ group is only moderately activating and mainly directs incoming electrophiles to the para position because the ortho positions are sterically hindered by the bulky $$-NHCOCH_3$$.

Step 3 : Controlled bromination
Bromine ($$Br_2$$) now reacts with acetanilide to give almost exclusively p-bromoacetanilide: $$C_6H_4Br(NHCOCH_3)$$.

Step 4 : Hydrolysis of the protecting group
Heating with water ($$H_2O,\,\Delta$$) hydrolyses p-bromoacetanilide back to p-bromoanilinium ion, releasing acetic acid: $$C_6H_4BrNH_3^+$$.

Step 5 : Liberation of free amine
Finally, aqueous $$NaOH$$ neutralises the anilinium ion and furnishes the free base, p-bromoaniline ($$C_6H_4BrNH_2$$).

Thus the only sequence that carries out protection → bromination → deprotection in the correct order is:

$$H_2SO_4,\; Ac_2O,\; Br_2,\; H_2O(\Delta),\; NaOH$$.

Option B is therefore the correct answer.