The problem describes a Dyson Sphere, a hypothetical structure completely enclosing the sun. We are given the radius of the sphere as $1.5 \times 10^8$ km and the thickness of the sphere as $3$ m.

1. Problem Description

The problem describes a Dyson Sphere, a hypothetical structure completely enclosing the sun. We are given the radius of the sphere as $1.5 \times 10^8$ km and the thickness of the sphere as $3$ m.

2. Solution Steps

Since the question does not specify any mathematical question, it is impossible to provide a specific solution. The image only provides descriptive information about the Dyson sphere concept and its dimensions. However, we can calculate the surface area of the sphere and the volume of the sphere shell.

The surface area $A$ of a sphere is given by the formula:

$A = 4\pi r^2$

where $r$ is the radius.

The volume $V$ of a spherical shell with outer radius $r_{outer}$ and inner radius $r_{inner}$ is given by:

$V = \frac{4}{3}\pi (r_{outer}^3 - r_{inner}^3)$

We are given the radius $r = 1.5 \times 10^8$ km. First, we convert this to meters:

$r = 1.5 \times 10^8 \text{ km} \times 1000 \frac{\text{m}}{\text{km}} = 1.5 \times 10^{11} \text{ m}$.

The thickness of the Dyson sphere is $3$ m. So, the outer radius is $r_{outer} = 1.5 \times 10^{11} + 3$ m, and the inner radius is $r_{inner} = 1.5 \times 10^{11}$ m.

Now we can approximate the volume since the thickness is small.

The volume $V$ of the Dyson sphere can be approximated as the surface area times the thickness $t$:

$V \approx A t = 4\pi r^2 t$

$V \approx 4 \pi (1.5 \times 10^{11})^2 (3) \approx 4 \pi (2.25 \times 10^{22}) (3) \approx 3.39 \times 10^{23} \times 4 \pi \times 3 = 8.48 \times 10^{23} m^3$

3. Final Answer

Since there is no specific mathematical question, a final answer cannot be given. However, based on the provided information, we calculated the volume of the Dyson sphere to be approximately $8.48 \times 10^{23} \text{ m}^3$. The surface area of the Dyson Sphere is $2.83 \times 10^{23} m^2$.