 Solution

Keep $b=t^{-1}at$. Repeatedly applying $ba{b}^{-1}=a^2$ givesIn the word metric on $G$ from the finite generating set $\{a,t\}$, the right-hand side therefore has length at most $6n+1$, because $b$ is represented by the word $t^{-1}at$. In the intrinsic word metric on the infinite cyclic group $⟨a⟩$ generated by $a$, however,A quasi-isometric embedding would bound the latter by an affine function of the former. The exponential sequence above violates every such bound, so the inclusion $⟨a⟩\hookrightarrow G$ is not a quasi-isometric embedding.