Semiconductor Electronics question for NEET / JEE Main, with solution

Answer: (D) $2$

One NAND gate gives $\overline{A\cdot B}$. A second NAND gate with its two inputs joined acts as a NOT gate and inverts this to $A\cdot B$. So two gates are enough, and one is not.

Solution by Sreeraj P, M.Sc Physics