In an aqueous solution containing sodium bicarbonate, aniline reacts quickly with bromine to give 2,4,6-tribromoaniline. nitration of aniline requires very strong conditions, however, and the yields (mostly m-nitroaniline) are poor.

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In bromination of aniline, due to the strongly activating NH2 group 2,4,6-tribromoaniline is formed.

In nitration of aniline the reagent used is concentrated nitric acid. Nitric acid reacts with aniline Nitrogen and protonates it. Due to NH3+ group formation, the benzene ring becomes strongly deactive in the ortho and para positions.

Hence the yields towards nitro nucleophile is poor.

The meta position being more electron rich than ortho or para due to the strongly deactivating group hence yields m-nitroaniline.

the answer is 75.0%.