Solution:

The region bounded by the parabola, $y^{2}=4 x$, and the line, $x=3$, is the area $\mathrm{OACO}$.

Area of OACO is symmetrical about $x$-axis.
$\therefore$ Area of $O A C O=2$ (Area of OAB)
$\text { Area OACO } \begin{aligned}
&=2\left[\int_{b}^{3} y d x\right] \\
&=2 \int_{0}^{3} 2 \sqrt{x} d x \\
&=4\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{3} \\
&=\frac{8}{3}\left[(3)^{\frac{3}{2}}\right] \\
&=8 \sqrt{3}
\end{aligned}$
As a result, the required area is $8 \sqrt{3}$ units.