A NAND gate is a universal gate. Use the standard constructions to find how many NAND gates are…

Step 1. NOT. One NAND gate with its inputs tied. $\overline{A \cdot A} = \overline{A}$.

Step 2. AND. One NAND gate, then one more as a NOT. That is $2$ gates.

Step 3. OR. A NOT on $A$ and a NOT on $B$ (two gates), then a NAND gate on the two results: $\overline{\overline{A} \cdot \overline{B}} = A + B$. That is $3$ gates.

Step 4. NOR. The OR circuit of $3$ gates, then one more as a NOT. That is $4$ gates.

Step 5. XOR. The four gates of the standard circuit. Gate 1 gives $\overline{AB}$. Gate 2 gives $\overline{A\,\overline{AB}}$, gate 3 gives $\overline{B\,\overline{AB}}$, and gate 4 combines them into $A \oplus B$.