The correct answer is $\boxed{\frac{16}{3}a^2}$.
The parabolas $y^2 = 4ax$ and $x^2 = 4ay$ intersect at the points $(\pm a, \pm a)$. The area between the parabolas is bounded by the lines $y = x$ and $y = -x$.
To find the area, we can use the following formula:
$$\text{Area} = \int_{-a}^a \left( x – y \right) \, dy$$
Substituting in the equations for the parabolas, we get:
$$\text{Area} = \int_{-a}^a \left( x – \sqrt{4ax} \right) \, dy$$
We can simplify this integral by using the substitution $u = \sqrt{4ax}$. This gives us:
$$\text{Area} = \int_{-2a}^2 \left( x – u \right) \, du$$
We can then evaluate this integral using the following formula:
$$\int u^n \, du = \frac{u^{n+1}}{n+1} + C$$
This gives us:
$$\text{Area} = \left[ \frac{x^2}{2} – \frac{u^2}{2} \right]_{-2a}^2 = \frac{4a^2}{2} – \frac{4a^2}{2} + \frac{4a^2}{2} – \frac{4a^2}{2} = \boxed{\frac{16}{3}a^2}$$