Yes, there is a $G$-structure of each finite order. In other words, for every $k\ge1$, there is an $n\ge1$ and a subgroup $G\subset GL(n,\mathbb{R})$ such that its Lie algebra $\frak{g}$ satisfies ${\frak{g}}^{(k-1)}\not=0$ while ${\frak{g}}^{(k)}=0$.
However, a theorem of Cartan (originally proved over $\mathbb{C}$ by a classification (but with some omissions), but later completed by others and shown to be valid over $\mathbb{R}$ as well) says that, if $G\subset GL(n,\mathbb{R})$ acts irreducibly on $\mathbb{R}^n$, then either $\frak{g}$ has order $1$, $2$, or $\infty$. (The list of the irreducibly acting $G\subset GL(n,\mathbb{R})$ that have order $2$ or $\infty$ is known.)