The first part of the exercise asked us to show that there's a normal, cyclic subgroup $H$ of order $85$. That's pretty easy using Sylow's third theorem, and generating a subgroup with elements of order $17$ and $5$.
Now, for the second part we have to show that $G\cong H\rtimes K$ with $K$ being a $2$-Sylow subgroup of $G$.
It'd also come in handy a hint on the third part, showing that $K$ can be $K\cong \mathbb Z_2\times \mathbb Z_2$ or $K\cong \mathbb Z_4$ if $G / |G|=340$ isn't abelian. Cheers!
So you have $H \lhd G$ with $|H|=85$. Let $K \in {\rm Syl}_2(G)$. Then $|K|=4$ and clearly $H \cap K = \{1\}$, so $|HK| = |H||K| = 340 = |G|$ and hence $HK=G$. So now $H$ and $K$ satisfy all the required conditions for an internal semidirect product, and we have $G = H \rtimes K$.
The two possibilities for $K$ are just the two isomorphism types of groups of order $4$.