Never ask a girl her age

Let $p$ be the age

Then $p=10x+y$ , where $x,y \in [0,9]$

and $p^2=100x^2+10xy+y^2$ $=10(10x^2+xy)+y^2$

since the last digit of $p^2$ (which is the last digit of $y^2$ ) equals to the first digit of $p$ (which is $x$ ) we find that $x \equiv y^2 \pmod{10}$ , and we can set this table:

and we can see that $p \in 0,11,42,93,64,55,66,97,48,19$

but since $p$ is a prime number, we find that $p \in 11,97,19$

$11^2=121$ , and 12 is not a perfect square

$97^2=9409$ , and 94 is not a perfect square

$19^2=361$ , and since 36 is a perfect square, the solution is $\boxed{19}$