The area of the region $$\{(x, y) : |x - y| \le y \le 4\sqrt{x}\}$$ is

We need to find the area of the region $$\{(x, y) : |x - y| \le y \le 4\sqrt{x}\}$$.

The condition $$|x - y| \le y$$ means $$-y \le x - y \le y$$, which gives us $$x \ge 0$$ (from the left inequality) and $$x \le 2y$$ (from the right inequality), i.e., $$y \ge x/2$$.

So the region is $$\{(x, y) : x \ge 0, \, x/2 \le y \le 4\sqrt{x}\}$$.

Finding the intersection of $$y = x/2$$ and $$y = 4\sqrt{x}$$: $$x/2 = 4\sqrt{x}$$, so $$x = 8\sqrt{x}$$, giving $$\sqrt{x}(\sqrt{x} - 8) = 0$$. Thus $$x = 0$$ or $$x = 64$$.

The area is:

$$A = \displaystyle\int_0^{64} \left(4\sqrt{x} - \dfrac{x}{2}\right) dx$$

$$= \left[4 \cdot \dfrac{2}{3}x^{3/2} - \dfrac{x^2}{4}\right]_0^{64}$$

$$= \dfrac{8}{3}(64)^{3/2} - \dfrac{64^2}{4}$$

$$= \dfrac{8}{3} \cdot 512 - \dfrac{4096}{4}$$

$$= \dfrac{4096}{3} - 1024 = \dfrac{4096 - 3072}{3} = \dfrac{1024}{3}$$

Hence, the correct answer is Option B.