Last is first

The last three digits of $3^{1997}$ are given by $3^{1997}\pmod{100}$ .

Since $3^{2}=1\pmod{4}$ , $3^{1997}=(3^2)^{998}3=3\pmod{4}$ . In particular, $3^{1997}=63\pmod{4}$ .

The totient $\phi(25)=20$ , so by Euler's Totient Theorem, $3^{20}=1\pmod{25}$ . Then

$3^{1997}=(3^{20})^{99}3^{17}=27^{5}\cdot3^2$

$=2^5\cdot9=13\pmod{25}$ .

In particular, $3^{1997}=63\pmod{25}$ .

Then since $4,25$ are coprime, $3^{1997}=63\pmod{4\cdot25}$ . That is, $3^{1997}=63\pmod{100}$ , so the answer is $\boxed{63}$ .