The product is supposed to be N,N-dimethylaniline.
This is simple. But if you are reading the textbook, you'll realize this is a bad question. Why?
The question proposer forgets, once you get aminobenzene, you will also get all the ortho and para methyl substituted byproducts that will form since, the amine group in aniline can (without LA catalysis) undergo Friedel-Crafts alkylation. It is extremely activated.
N-alkylation is inhibited by disruption of conjugation. You typically cannot just alkylate nitrogens in amines without causing exhaustive methylation. You will have trimethylaniiline too. But you will create a lot of undesired ortho and para toluidine (aminotoluene). And the reagents as presented will mean no single product for the second step.
The first step is nitration where sulfuric acid and nitric acid exchange protons and the nitronium electrophile is produced. http://bit.ly/1Rj1PRO Benzene attacks nitronium in the rate determining step and then undergoes an E1 type elimination to restore aromaticity to the ring giving nitrobenzene. Nitrobenzene is reduced by Sn at reflux in HCl to give arylammonium chloride salts. The workup which is actually another procedure (stoichiometric amounts base are required) and is left out entirely and required to produce aniline.
With just arylammonium cations the next step is no reaction (another mistake in the question). You cannot alkylate an already formally positive amine, that is certain. And doing so (even with aniline) by adding 2 equivalents will lead to unintended mixtures of toluidine and N,N,N-trimethylanilinium iodide, N,N-dimethylaniline, N-methylaniline, and just unreacted aniline. The HI created will also protonate the aniline further making this reaction unfeasible.
Students are told so many white lies but as many white lies as the enormous number of glaring omissions in undergraduate lecture and organic chemistry textbooks. Presumably it's believed necessary to simplify things but that just makes all the people who think critically totally confused.