In order that an upper triangular matrix
$$\begin{pmatrix} a & b & c \\ 0 & d& e \\ 0 &0 & f\end{pmatrix}$$
be invertible, it is necessarily and sufficient that the determinant $adf$ be nonzero. This means that $b,c,e$ can be anything you want, while $a, d, f$ cannot be zero. This means there are
$$3 \cdot 3 \cdot 3 \cdot 2 \cdot 2 \cdot 2 = 216 $$
such matrices.
The subgroup $D$ of diagonal invertible matrices is as far from being normal in $G$ as you can get (if $x \in G$, but not in $D$, then $xDx^{-1} \neq D$).
However, if you multiply two upper triangular matrices $x, y \in G$, notice that the entries on the diagonal of $xy$ are obtained by multiplying the corresponding entries on the diagonal of $x$ and $y$. This implies that
$$N = \{ \begin{pmatrix} 1 & b & c \\ 0 & 1& e \\ 0 &0 & 1\end{pmatrix} : b, c, e \in \mathbb{F}_3 \}$$ is a normal subgroup of $G$.
Actually, $G$ is the semidirect product of $N$ and $D$.
Best Answer
Hint: Realize $\mathbb{F}_{q^n}^*$ as a subgroup of $\mathrm{GL}_n(\mathbb{F}_q)$.