Astrophysicists have discovered a minor planet of radius $30$ kilometers whose surface is completely covered in water. A spherical meteor hits this planet and is submerged in the water. This incidence causes an increase of $1$ centimeters to the height of the water on this planet. What is the radius of the meteor in meters?
1. **Determine the volume of water displaced by the meteor:**
   The increase in water height is \(1\) centimeter, which is \(0.01\) meters. The radius of the planet is \(30\) kilometers, which is \(30,000\) meters. The volume of water displaced, \(V\), can be calculated using the formula for the volume of a spherical shell:
   \[
   V = 4\pi R^2 h
   \]
   where \(R\) is the radius of the planet and \(h\) is the increase in water height.
   \[
   V = 4\pi (30,000)^2 \times 0.01
   \]
   \[
   V = 4\pi \times 900,000,000 \times 0.01
   \]
   \[
   V = 36,000,000\pi \text{ cubic meters}
   \]

2. **Relate the volume of the meteor to the volume of water displaced:**
   The volume of the meteor, \(V_m\), is equal to the volume of water displaced:
   \[
   V_m = \frac{4}{3}\pi r^3
   \]
   where \(r\) is the radius of the meteor. Setting this equal to the volume of water displaced:
   \[
   \frac{4}{3}\pi r^3 = 36,000,000\pi
   \]

3. **Solve for the radius of the meteor:**
   \[
   \frac{4}{3} r^3 = 36,000,000
   \]
   \[
   r^3 = 36,000,000 \times \frac{3}{4}
   \]
   \[
   r^3 = 27,000,000
   \]
   \[
   r = \sqrt[3]{27,000,000}
   \]
   \[
   r = 300 \text{ meters}
   \]

The final answer is \(\boxed{300 \text{ meters}}\)