Answer
$\text {The vertical line that divides the area in equal two parts is}$
\begin{align}
x = \sqrt[3] 4
\end{align}
Work Step by Step
$\text {First, we have to find the area between x = $\sqrt y$, x = 2, x = 0.}$
$\text {The intersection between the functions:}$
\begin{align}
2 = \sqrt y \\
y = 4
\end{align}
$\text {Thus, the area is}$
\begin{align}
A = \int_0^4 2 \ dy - \int_0^4 \sqrt y \ dy = 8 -\left[\frac{2}{3} y^{\frac{3}{2}} \right]_0^4 = 8 - \frac{16}{3} = \frac{8}{3}
\end{align}
$\text {The half of the area is $\frac{4}{3}$. Now we will use it to find x = k as follows}$
$\text {The intersection between the functions:}$
\begin{align}
k = \sqrt y \\
y = k^2
\end{align}
$\text{Thus, the area is}$
\begin{align}
A = \frac{4}{3} = \int_0^{k^2} k \ dy - \int_0^{k^2} \sqrt y \ & dy = k^3 - \left[\frac{2}{3} y^{\frac{3}{2}} \right]_0^{k^2} = k^3 - \frac{2}{3}k^3 = \frac{4}{3} \Rrightarrow \\
\Rrightarrow k = \sqrt[3] 4
\end{align}
$\text {Finally, the vertical line that divides the area in equal two parts is}$
\begin{align}
x = \sqrt[3] 4
\end{align}
