Nitrobenzene undergoes reduction with $\mathrm{Zn}$ aqueous $\mathrm{KOH}$ to form a compound ' $\mathrm{A}$ '. The number of sigma and Pi bonds in ' $\mathrm{A}$ ' respectively are 

Nitrobenzene in the presence of $\mathrm{Zn}$ and an aqueous medium $\mathrm{KOH}$ gets reduced to form hydrazo benzene. This product hydrazo benzene is compound A. Two benzene rings attached to each other via N-N sigma bond. 

Sigma bonds denote the single bonds in the benzene ring and the hydrogen bonds attached to them along with the $H N-N H$bonds. And Pi bonds are the double bonds in the benzene ring.

Hence, the total number of sigma bonds is 27 and the number of Pi bonds is 6. 

Therefore, option (B) is correct.