HINT They ask for the area of the yellow region:
The areas would be given by integrals $\int_{x_1}^{x_2} \left(y_\text{top}(x) - y_\text{bottom}(x)\right) \mathrm{d} x$ with appropriate choices of boundaries $x_1$ and $x_2$ and functions $y_\text{top}(x)$ and $y_\text{bottom}(x)$.
We can find the area by integrating with respect to $x$ or with respect to $y$.
With respect to $y$ is easier. I would prefer to use symmetry and say that the area is equal to
$$2\int_0^4 y^{1/2}\,dy.$$
Integrate. We get $(2)(2/3)4^{3/2}=32/3$, your answer.
So we want the area from $y=0$ to $b$ to be half of $\frac{32}{3}$, that is, $\frac{16}{3}$.
The area from $y=0$ to $y=b$ is equal to
$$2\int_0^b y^{1/2}\,dy.$$
This is equal to $(2)(2/3)b^{3/2}$.
Thus we need to find the $b$ such that
$$(2)(2/3)b^{3/2}=\frac{16}{3}.$$
Simplify. We get the equation $b^{3/2}=4$.
Solve for $b$. We get $b=4^{2/3}$. This can be written in various other ways, such as $b=2\sqrt[3]{2}$.
Remark: We could have bypassed the initial computation of the area, and written that we want to find $b$ such that
$$2\int_0^b y^{1/2}\,dy=2\int_b^4 y^{1/2}\,dy.$$
But I think it is better to find first the area from $0$ to $4$, to get some concrete feeling about the situation.
For a "reality check" note that our answer must be $\gt 2$: we have to go more than half of the way to $4$ to capture half the area.
Best Answer
The curves $y = x^2$ and $y=x^3$ intersect a x=0,1 for x>=0
If we find the values of $x^2$ and $x^3$ for $0<=x<=1$, we can find that $x^2>=x^3$
Now for the double integral,$\int\int_D dA = \int\int_D dxdy$, we have the limits as, $0<=x<=1$ and $x^3<=y<=x^2$
So,
$\int\int_D dA = \int^1_{0}\int^{x^2}_{x^3} dxdy = \int^1_{0} dx(y)^{x^2}_{x^3} = \int^1_{0} (x^2 - x^3)dx$ $$= (\frac{x^3}{3} - \frac{x^4}{4})^1_0= (\frac{1}{3} - \frac{1}{4}) = \frac{1}{12}$$
Now, $I = 2\int\int_D dA = 2.\frac{1}{12} = \frac{1}{6}$