This is probably totally wrong.
We know that $5$ and $63$ are relatively prime, therefore $w^5 \in G_{63}$ primitive (we'll suppose $w= w^5$ without loss of generality).
We also know that $(w^7)^9 = w^{63} = 1$, so $w^7 \in G_9$.
If $w^7$ were primitive (is it?, how do I prove this?), then
$$\sum_{k=6}^n w^{35k} = \sum_{k=6}^n (w^7)^k = 0 \ \iff \ n \equiv 5 \pmod 9$$
Because we need the sum of terms $\{6,7,\dots,n\}$ to span a complete set of residues (modulo $9$) a fixed number of times.
On the other hand,
$$w^{12n} = w^{15} \ \iff \ (w^3)^{4n} = (w^3)^5 \ \iff \ n \equiv 17 \pmod {21}$$
So the final answer is $n \equiv 59 \pmod {63}$?