The area of the region enclosed between the curves 4x² = y and y = 4 is:

The correct option is 16/3 sq. units.

To find the area of the region enclosed between the curves, we first identify the equations of the given curves:
1. A parabola: ${\mathrm{4x}}^2=y$, which can be rewritten as $x^2=\frac{y}{4}$ or $x=\pm \frac{\sqrt{y}}{2}$.
2. A horizontal line: $y=4$.

The region is bounded below by the parabola $y={\mathrm{4x}}^2$ and above by the line $y=4$.
To find the points of intersection, we set the two equations equal to each other:
${\mathrm{4x}}^2=4$
$x^2=1$
$x=\pm 1$

Thus, the curves intersect at the points $(-1,4)$ and $(1,4)$. The region is symmetric with respect to the y-axis.

We can calculate the area, A, by integrating with respect to x from $x=-1$ to $x=1$:
$A={\int }_{-1}^1(4-{\mathrm{4x}}^2)dx$

Due to the symmetry about the y-axis, we can also write this as:
$A=2{\int }_0^1(4-{\mathrm{4x}}^2)dx$

Let's evaluate this definite integral step-by-step:
$A=2{[\mathrm{4x}-\frac{{\mathrm{4x}}^3}{3}]}_0^1$
$A=2\left[\right(4\left(1\right)-\frac{4{\left(1\right)}^3}{3})-0]$
$A=2[4-\frac{4}{3}]$
$A=2\left[\frac{12-4}{3}\right]$
$A=2\left[\frac{8}{3}\right]$
$A=\frac{16}{3}$

Therefore, the area of the enclosed region is indeed 16/3 sq. units.