To solve this problem, let's analyze the circuit. We have:

1. A constant voltage \( U = 18 \, \text{V} \).
2. Internal resistance \( r = 2 \, \Omega \).
3. A light bulb \( Đ \) with a rated voltage of 6 V.
4. An adjustable resistance \( R \).

The current through the ammeter is 1 A, which means the total current in the circuit is 1 A.

### Given:
- Total voltage \( U = 18 \, \text{V} \)
- Internal resistance of battery \( r = 2 \, \Omega \)
- Current \( I = 1 \, \text{A} \)
- Rated voltage of the bulb \( V_{\text{bulb}} = 6 \, \text{V} \)

### Let's find:
- The rated power of the bulb \( P_{\text{bulb}} \).

### Steps:

1. **Voltage across Bulb**:
Since the bulb is operating normally, it uses its rated voltage. Therefore, the voltage across the bulb is \( V_{\text{bulb}} = 6 \, \text{V} \).

2. **Voltage across the Internal Resistance**:
Ohm's Law: \( V_r = I \times r = 1 \, \text{A} \times 2 \, \Omega = 2 \, \text{V} \).

3. **Voltage across the Resistor \( R \)**:
Total voltage \( U = 18 \, \text{V} \).
The sum of voltages across all components in the circuit should be equal to the total voltage:
\[
V = V_{\text{bulb}} + V_r + V_R
\]
\[
18 \, \text{V} = 6 \, \text{V} + 2 \, \text{V} + V_R
\]
\[
V_R = 18 \, \text{V} - 6 \, \text{V} - 2 \, \text{V} = 10 \, \text{V}
\]

4. **Rated Power of the Bulb**:
The power \( P \) of the bulb can be calculated using the formula \( P = V \times I \):
\[
P_{\text{bulb}} = V_{\text{bulb}} \times I = 6 \, \text{V} \times 1 \, \text{A} = 6 \, \text{W}
\]

So, the rated power of the bulb is 6 Watts.