Hence the area of the region bounded by the curve $y^2=4x$,y-axis and the line y=3 is the shaded portion as shown in the fig.
Hence $A=\int_0^3x\;dy.$
Here $x=\frac{y^2}{4}$
$A=\int_0^3\frac{y^2}{4}\;dy.$
On integrating we get,
$A=\frac{1}{4}\begin{bmatrix}\frac{y^3}{3}\end{bmatrix}_0^3$.
On applying limits we get,
$A=\frac{1}{12}[3^3-0]$
$\;\;\;\;=\frac{1}{12}\times 27=\frac{9}{4}$sq.units.
Hence B is the correct answer.