The product of oxidation of aniline with ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ and $\mathrm{conc}$. ${\mathrm{H}}_2{\mathrm{SO}}_4$ will be. 

Potassium dichromate, ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ is known to be a very strong oxidizing agent that readily tends to add oxygen to the compound. And aniline is an aromatic primary amine undergoes oxidation in the presence of potassium dichromate and concentrated sulphuric acid to form 1,4 - benzoquinone along with the removal of an ammonium ion.

This reaction happens in an acidic medium. So, the product formed when aniline is treated with ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ and conc. ${\mathrm{H}}_2{\mathrm{SO}}_4$ is p-benzoquinone.

Therefore, option (B) is correct.