To solve the problem, we need to find the amplitude of the electric field (E₀) of an electromagnetic wave emitted by a point source. Given the power emitted by the source and the distance from the source, we can use the relationship between power, intensity, and the electric field.
### Step-by-Step Solution:
1. **Understand the relationship between power, intensity, and electric field**:
The average intensity (I) of an electromagnetic wave can be expressed in two ways:
- \( I = \frac{P}{A} \) where \( A \) is the area over which the power is distributed.
- \( I = \frac{1}{2} \epsilon_0 E_0^2 c \) where \( E_0 \) is the amplitude of the electric field, \( \epsilon_0 \) is the permittivity of free space, and \( c \) is the speed of light.
2. **Calculate the area at a distance r**:
For a point source, the area \( A \) at a distance \( r \) is given by:
\[
A = 4 \pi r^2
\]
Here, \( r = 15 \, \text{km} = 15 \times 10^3 \, \text{m} \).
3. **Substitute the values into the intensity formula**:
The intensity can be expressed as:
\[
I = \frac{P}{4 \pi r^2}
\]
Substituting \( P = 15 \, \text{W} \) and \( r = 15 \times 10^3 \, \text{m} \):
\[
I = \frac{15}{4 \pi (15 \times 10^3)^2}
\]
4. **Calculate the intensity**:
First, calculate \( 4 \pi (15 \times 10^3)^2 \):
\[
4 \pi (15 \times 10^3)^2 = 4 \pi \times 225 \times 10^6 = 900 \pi \times 10^6
\]
Therefore, the intensity \( I \) becomes:
\[
I = \frac{15}{900 \pi \times 10^6}
\]
5. **Equate the two expressions for intensity**:
Now, using the second expression for intensity:
\[
I = \frac{1}{2} \epsilon_0 E_0^2 c
\]
We know \( \frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \) and \( c = 3 \times 10^8 \, \text{m/s} \).
6. **Rearranging for \( E_0 \)**:
From the intensity equations, we can set them equal:
\[
\frac{15}{900 \pi \times 10^6} = \frac{1}{2} \epsilon_0 E_0^2 c
\]
Rearranging gives:
\[
E_0^2 = \frac{2 \times 15}{900 \pi \times 10^6 \times \epsilon_0 \times c}
\]
7. **Substituting values**:
Substitute \( \epsilon_0 = \frac{1}{4 \pi (9 \times 10^9)} \):
\[
E_0^2 = \frac{30}{900 \pi \times 10^6 \times \frac{1}{4 \pi \times 9 \times 10^9} \times 3 \times 10^8}
\]
8. **Simplifying**:
After substituting and simplifying, we find:
\[
E_0 = \sqrt{\frac{30 \times 9 \times 10^9}{3 \times 225 \times 10^{14}}}
\]
9. **Final Calculation**:
After calculating, we find:
\[
E_0 = 2 \times 10^{-3} \, \text{V/m}
\]
### Conclusion:
Thus, the value of \( x \) in \( E_0 = x \times 10^{-3} \, \text{V/m} \) is:
\[
\boxed{2}
\]
