$\Delta=\frac{1}{2}\left|\begin{array}{lll}x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1\end{array}\right|$

It is given that the area of triangle is $4$ square units.

$\therefore \Delta=\pm 4$

The area of the triangle with vertices $(-2,0),(0,4),(0, k)$ is given by the relation,

$\Delta=\frac{1}{2}\left|\begin{array}{ccc}-2 & 0 & 1 \\ 0 & 4 & 1 \\ 0 & k & 1\end{array}\right|$

$=\frac{1}{2}[-2(4-k)]$

$=k-4$

$\therefore k-4=\pm 4$

When $k-4=-4, k=0$

When $k-4=4, k=8$

Hence, $k=0,8$