3.
\(\;\; \begin{array} \hline \;\;\;1 & A\\ \;\;×& A\\ \hline \;\;9 & A\\ \hline \end{array} \)

We can observe that A×A gives A as the unit's digit to the multiplication, that means it's only possible if A=1, 6 and 5.

Checking for all the case:

For A=1: \( \;\;\;\begin{array} \hline \;\;\;1 & 1\\ \;\;×& 1\\ \hline \;\;1 & 1\\ \hline \end{array} \)

We can see clearly that it is not similar with the given multiplication so A\(\neq\)1.

For A=5: \( \;\;\;\begin{array} \hline \;\;\;1 & 5\\ \;\;×& 5\\ \hline \;\;7 & 5\\ \hline \end{array} \)

We can see clearly that it is not similar with the given multiplication so A\(\neq\)5.

For A=6: \( \;\;\;\begin{array} \hline \;\;\;1 & 6\\ \;\;×& 6\\ \hline \;\;9 & 6\\ \hline \end{array} \)

We can see clearly that it is similar with the given multiplication so A=6.