Since, given parabola is symmetric about the $y-$ axis, hence rectangle will also be symmetric about $y-$ axis.

Let one vertex of the rectangle on the $x-$ axis be $\left(\alpha ,0\right)$, then

Area of rectangle $A=2\alpha .\left(12-{\alpha }^2\right)$

Differentiating both sides with respect to $\alpha$, we get

$\Rightarrow \frac{dA}{d\alpha }=24-6{\alpha }^2=0\Rightarrow \alpha =2,-2$

For area to be maximum, put $\alpha =2$ in the equation for area of the rectangle, we get

$A_{max}=2\times 2\times \left(12-2^2\right)=32$