Step 1: Understand the region.
We want the area trapped between the rightward parabola $y^2=8x$ and the upward parabola $x^2=8y$. By symmetry they cross at the origin and at one more point.
Step 2: Find the intersection points.
From $x^2=8y$ we get $y=\dfrac{x^2}{8}$. Put this into $y^2=8x$: $\left(\dfrac{x^2}{8}\right)^2=8x$, so $\dfrac{x^4}{64}=8x$, giving $x^4=512x$ and $x(x^3-512)=0$. Thus $x=0$ or $x=8$.
Step 3: Identify upper and lower curves.
Between $x=0$ and $x=8$, the top boundary is $y=\sqrt{8x}=2\sqrt{2}\,\sqrt{x}$ (from $y^2=8x$) and the bottom boundary is $y=\dfrac{x^2}{8}$ (from $x^2=8y$).
Step 4: Set up the area integral.
$\text{Area}=\displaystyle\int_{0}^{8}\left(2\sqrt{2}\,\sqrt{x}-\dfrac{x^2}{8}\right)dx$.
Step 5: Integrate term by term.
$\displaystyle\int_{0}^{8}2\sqrt{2}\,x^{1/2}\,dx=2\sqrt{2}\cdot\dfrac{2}{3}x^{3/2}\Big|_0^8=\dfrac{4\sqrt{2}}{3}(8)^{3/2}=\dfrac{4\sqrt{2}}{3}\cdot16\sqrt{2}=\dfrac{128\cdot2}{3}=\dfrac{256}{3}$, and $\displaystyle\int_{0}^{8}\dfrac{x^2}{8}\,dx=\dfrac{1}{8}\cdot\dfrac{x^3}{3}\Big|_0^8=\dfrac{512}{24}=\dfrac{64}{3}$.
Step 6: Subtract to get the answer.
$\text{Area}=\dfrac{256}{3}-\dfrac{64}{3}=\dfrac{192}{3}\cdot\dfrac{1}{1}$... carefully, $\dfrac{256-64}{3}=\dfrac{192}{3}$? Re-evaluate the first integral: $(8)^{3/2}=22.63=16\sqrt2$, so first term $=\dfrac{4\sqrt2}{3}\cdot16\sqrt2=\dfrac{4\cdot16\cdot2}{3}=\dfrac{128}{3}$. Then $\text{Area}=\dfrac{128}{3}-\dfrac{64}{3}=\dfrac{64}{3}$ sq. units, matching option (2).
\[ \boxed{\dfrac{64}{3}\ \text{sq. units}} \]