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

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

$\therefore \Delta=\pm 4$

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

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

$=\frac{1}{2}[k(0-2)-0(4-0)+1(8-0)]$

$=\frac{1}{2}[-2 k+8]=k+4$

$\therefore-k+4=\pm 4$

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

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

Hence, $k=0,8$