$a^2 \, y = x^3$
when $x = 0, \,\, y = 0$
when $x = 2a, \,\, y = 8a$
points of intersection: (0, 0) and (2a, 8a)
Using vertical strip:
$\displaystyle A = {\int_{x_1}}^{x_2} y \, dx$
$\displaystyle A = {\int_0}^{2a} \dfrac{x^3 \, dx}{a^2}$
$A = \left[ \dfrac{x^4}{4a^2} \right]_0^{2a}$
$A = \dfrac{(2a)^4 - 0^4}{4a^2}$
$A = 4a^2 \,\, \text{unit}^2$ answer