Interpret the region

The set is intended to represent the region bounded by the parabola $y^2=4x,$ the vertical lines $x=0$ and $x=3$.

So the region is $\{(x,y): y^2\le 4x,\; 0\le x\le 3\}.$

Find the vertical extent for a fixed $x$

From $y^2\le 4x,$ we get $-2\sqrt{x}\le y\le 2\sqrt{x}.$

Hence, for each $x\in[0,3]$, the vertical length of the strip is $2\sqrt{x}-(-2\sqrt{x})=4\sqrt{x}.$

Set up the area integral

Therefore, the required area is $A=\int_0^3 4\sqrt{x}\,dx.$

Evaluate the integral

Compare with the options

The computed area is $\boxed{8\sqrt{3}},$ which is not among the given options $\frac{32}{3}, \frac{16}{3}, \frac{8}{3}, \frac{64}{3}$.

About the stored answer

The stored correct answer is $\frac{32}{3}$, but that would correspond to a different parabola, for example if the equation were $y^2=16x,$ then the area from $x=0$ to $x=3$ would be $\int_0^3 8\sqrt{x}\,dx=16\sqrt{3},$ still not $\frac{32}{3}$.

If instead the intended curve were $x=\frac{y^2}{4},\quad 0\le y\le 4,$ then the area could become $\frac{16}{3}$ or similar depending on bounds. So the printed question likely has a typo or formatting issue.

For the region as written/interpreted, the correct area is $\boxed{8\sqrt{3}}.$