To find the ratio of power outputs when two heaters with different power ratings are connected first in series and then in parallel, we need to understand how the total power output varies based on the type of connection.

Heater Specifications:

• Power of heater A $$ (P_A) = 1 \, \text{kW} = 1000 \, \text{W} $$
• Power of heater B $$ (P_B) = 2 \, \text{kW} = 2000 \, \text{W} $$

Scenario 1: Series Connection

When resistors (or heaters in this case) are connected in series, the total resistance ($R_{\text{series}}$) is the sum of the individual resistances ($R_A$ and $R_B$).

Using the formula for electrical power: $ P = \frac{V^2}{R} $,

where $ P $ is power, $ V $ is voltage, and $ R $ is resistance, we can express the resistance of each heater as:

$$ R_A = \frac{V^2}{P_A} $$

$$ R_B = \frac{V^2}{P_B} $$

Substitute the given power values:

$$ R_A = \frac{V^2}{1000} $$

$$ R_B = \frac{V^2}{2000} $$

Then the total resistance for the series connection is:

$$ R_{\text{series}} = R_A + R_B = \frac{V^2}{1000} + \frac{V^2}{2000} = \frac{3V^2}{2000} $$

The total power output in series ($P_{\text{series}}$) is:

$$ P_{\text{series}} = \frac{V^2}{R_{\text{series}}} = \frac{V^2}{\frac{3V^2}{2000}} = \frac{2000}{3} \text{W} $$

Scenario 2: Parallel Connection

For parallel connections, the total resistance ($R_{\text{parallel}}$) is given by:

$$ \frac{1}{R_{\text{parallel}}} = \frac{1}{R_A} + \frac{1}{R_B} = \frac{1}{\frac{V^2}{1000}} + \frac{1}{\frac{V^2}{2000}} = \frac{3}{2V^2} $$

Reformulate to find $R_{\text{parallel}}$:

$$ R_{\text{parallel}} = \frac{2V^2}{3} $$

And the total power output in parallel ($P_{\text{parallel}}$) is:

$$ P_{\text{parallel}} = \frac{V^2}{R_{\text{parallel}}} = \frac{V^2}{\frac{2V^2}{3}} = \frac{3V^2}{2} $$

However, simplifying,

$$ P_{\text{parallel}} = \frac{3}{2} V^2 $$

The ratio of powers is then:

$$ \frac{P_{\text{series}}}{P_{\text{parallel}}} = \frac{\frac{2000}{3}}{\frac{3V^2}{2}} $$

Solving and simplifying,

$$ \text{Ratio} = \frac{\frac{2000}{3}}{\frac{3 \times V^2}{2}} = \frac{2000 \times 2}{3 \times 3 \times V^2} = \frac{4000}{9V^2} $$

Given that we know one ratio of the actual power values, we simplify further. For calculating power in simple terms, consider voltage to be normalized (taken out of the fraction):

$$ \frac{P_{\text{series}}}{P_{\text{parallel}}} = \frac{\frac{2000}{3}}{\frac{6000}{2}} = \frac{2000 \times 2}{3 \times 6000} = \frac{4000}{18000} = \frac{2}{9} $$

Therefore, the ratio of the power outputs when the heaters are connected first in series and then in parallel is 2:9, which corresponds to Option B.