Clearly the equation $y^2=4x$ represents parabola with vertex (0,0) at the origin and axis of symmetry along the positive direction of x-axis as shown in the fig.
x= 3 is a line parallel to the y-axis.The region is the shaded portion as shown in the fig.
Since $y^2=4x is symmetrical about x-axis the required area is
A=2(area bounded by the parabola and the line and the x-axis).
Hence the required area is given by
$A=2\int_3^0y\;dx=2\int_3^02\sqrt x\;dx.$
$A=4\int_3^0x^\frac{1}{2}dx.$
on integrating we get,
$A=4\begin{bmatrix}\frac{x^\frac{3}{2}}{\frac{3}{2}}\end{bmatrix}_0^3$
$\;\;\;=4\times\frac{2}{3}[x^\frac{3}{2}]_0^3$
on applying limits we get,
$A=\frac{8}{3}\begin{bmatrix}(3)^\frac{3}{2}\end{bmatrix}$sq.units.
$\;\;\;=\frac{8}{3}\times 3\sqrt 3$
$\;\;\;=8\sqrt 3sq.units.$
Hence the required area is $8\sqrt 3$sq.units.