The length of each cross-section is determined by the horizontal distance (parallel to the x-axis) from one end of the parabola to the other.
[tex]y=4-2x^2\implies x=\pm\sqrt{\dfrac{4-y}2}[/tex]
which means this horizontal distance is
[tex]\sqrt{\dfrac{4-y}2}-\left(-\sqrt{\dfrac{4-y}2}\right)=2\sqrt{\dfrac{4-y}2}[/tex]
The area of each cross-section is simply the square of the section's side length, so the area would be [tex]\left(2\sqrt{\dfrac{4-y}2}\right)^2=4\dfrac{4-y}2=8-2y[/tex].
So the volume of the solid would be
[tex]\displaystyle\int_0^4(8-2y)\,\mathrm dy=16[/tex]
