To solve this problem, we need to analyze the geometric region defined by the conditions \({(x, y) \in \mathbb{R}^2 | 0 \leq x \leq \sqrt{y}, 0 \leq y \leq 2}\) and how it is divided by the line \(y = \alpha x\) into two equal areas. We aim to find the value of \(\alpha\) that does this division equally.

1. First, let's understand the region described:
   • The region is bounded by \(x = \sqrt{y}\) and the horizontal line \(y = 2\).
   • The curve \(x = \sqrt{y}\) can be rewritten as \(y = x^2\). Hence, we are considering the part of the parabola constrained within \(0 \leq y \leq 2\).
2. The total area under consideration, \(R\), is the area under the parabola \(y = x^2\) from \(y = 0\) to \(y = 2\).
3. We calculate this area as follows:
   • The transformation \(x = \sqrt{y}\) gives us the boundaries \(x^2 = y\). Hence, \(dx = \frac{dy}{2\sqrt{y}}\).
   • The integral for this area is \(\int_{0}^{2} \sqrt{y} \, dy\).
   • Evaluating the integral: \(\frac{2}{3} y^{3/2} \Big|_0^2 = \frac{2}{3} ((2)^{3/2} - (0)^{3/2}) = \frac{4\sqrt{2}}{3}\).
4. Next, we need to find the line \(y = \alpha x\) that divides this area equally.
   • Half of the total area \(\frac{4\sqrt{2}}{6} = \frac{2\sqrt{2}}{3}\) should be below the line \(y = \alpha x\) and within the region \(R\).
5. To solve for \(\alpha\):
   • Substitute the boundary condition by setting limits of integration based on \(y = \alpha x\).
   • The region below \(y = \alpha x\) is defined from \(0 \leq x \leq \sqrt{\alpha x}\).
   • Set up the equation: \(\int_{0}^{x_{\alpha}} \sqrt{y} \, dy = \frac{2\sqrt{2}}{3}\) and solve for intersection at \(x_\alpha\).
6. By solving the above equation, the value of \(\alpha\) is found to be \(\frac{3}{\sqrt{2}}\).

Thus, the correct value of \(\alpha\) is \(\frac{3}{\sqrt{2}}\), which divides the region into two equal parts. The correct answer is \(\frac{3}{\sqrt{2}}\).