We are given the region $R$ bounded by the parabola $y = \frac{1}{16}x^2$ and the line $y = 5$. The volume of the solid is formed by cross-sections perpendicular to the $x$-axis, which are rectangles where the height is twice the side in the plane of the base.

Step 1: Determine the bounds for $x$

To find the intersection points between $y = \frac{1}{16}x^2$ and $y = 5$, set the two equations equal to each other:

$\frac{1}{16}x^2 = 5$

Multiply both sides by 16:

$x^2 = 80$

Taking the square root:

$x = \pm \sqrt{80} = \pm 4\sqrt{5}$

Thus, the region $R$ is bounded by $x = -4\sqrt{5}$ and $x = 4\sqrt{5}$.

Step 2: Cross-sectional area of rectangles

For each cross-section at a given $x$, the base of the rectangle is the vertical distance between the parabola $y = \frac{1}{16}x^2$ and the line $y = 5$. This distance (or the side of the rectangle in the plane of the base) is:

$\text{side} = 5 - \frac{1}{16}x^2$

Since the height of the rectangle is twice the side in the base:

$\text{height} = 2 \times \left( 5 - \frac{1}{16}x^2 \right)$

Thus, the area $A(x)$ of each cross-sectional rectangle is:

$A(x) = \text{side} \times \text{height} = \left( 5 - \frac{1}{16}x^2 \right) \times 2\left( 5 - \frac{1}{16}x^2 \right)$

$A(x) = 2 \left( 5 - \frac{1}{16}x^2 \right)^2$

Step 3: Volume of the solid

The volume $V$ of the solid is the integral of the cross-sectional area along the $x$-axis from $x = -4\sqrt{5}$ to $x = 4\sqrt{5}$:

$V = \int_{-4\sqrt{5}}^{4\sqrt{5}} A(x) \, dx = 2 \int_{-4\sqrt{5}}^{4\sqrt{5}} \left( 5 - \frac{1}{16}x^2 \right)^2 dx$

Since the integrand is an even function (it depends on $x^2$), we can simplify the limits of integration by doubling the integral from 0 to $4\sqrt{5}$:

$V = 4 \int_0^{4\sqrt{5}} \left( 5 - \frac{1}{16}x^2 \right)^2 dx$

Step 4: Expand and integrate

First, expand $\left( 5 - \frac{1}{16}x^2 \right)^2$:

$\left( 5 - \frac{1}{16}x^2 \right)^2 = 25 - 2 \times 5 \times \frac{1}{16}x^2 + \left( \frac{1}{16}x^2 \right)^2$

$= 25 - \frac{10}{16}x^2 + \frac{1}{256}x^4$

Now, integrate term by term:

$V = 4 \int_0^{4\sqrt{5}} \left( 25 - \frac{10}{16}x^2 + \frac{1}{256}x^4 \right) dx$

$= 4 \left[ 25x - \frac{10}{16} \frac{x^3}{3} + \frac{1}{256} \frac{x^5}{5} \right]_0^{4\sqrt{5}}$

Step 5: Evaluate the integral at $x = 4\sqrt{5}$

Substitute $x = 4\sqrt{5}$ into each term:

1. $25x = 25(4\sqrt{5}) = 100\sqrt{5}$

2. $\frac{10}{16} \frac{x^3}{3} = \frac{5}{8} \times \frac{(4\sqrt{5})^3}{3} = \frac{5}{8} \times \frac{64 \times 5^{3/2}}{3} = \frac{5 \times 320\sqrt{5}}{24} = \frac{1600\sqrt{5}}{24} = \frac{400\sqrt{5}}{6}$

3. $\frac{1}{256} \frac{x^5}{5} = \frac{1}{256} \times \frac{(4\sqrt{5})^5}{5} = \frac{1}{256} \times \frac{1024 \times 5^{5/2}}{5} = \frac{1}{256} \times 2048 \times 5^{3/2} = \frac{2048\sqrt{5}}{1280}$

Now simplify and subtract.

Finally,