Its the first time I'm seeing such question, I tried something but I can't really confirm it since its pretty new to me:
Its easy to prove with sylow that there is only one subgroup $H$ with order $17^2$, hence, $H$ is normal. Since $17$ is prime, and $|H|=17^2$ then there are $2$ groups up to isomorphism with that order. $G/H$ is a group of order $5^2$ (since $H$ is normal), and again, we have $2$ groups with that order up to isomorphism.
So the answer is $2\times 2+1=5$? (Added one for the case when $G$ is cyclic). is there anything else I missed here?