To solve this problem, we will use the Stefan-Boltzmann law, which states that the power radiated by a blackbody is given by:$\\ P = \sigma A T^4 \\$where $\sigma$ is the Stefan-Boltzmann constant, $A$ is the surface area, and $T$ is the temperature in Kelvin.$\\$Given:$\\$• Initial radius $r_1 = 12$ cm$= 0.12$ m$\\$• Initial temperature $T_1 = 500$ K$\\$• Initial power $P_1 = 450$ W$\\$The surface area of a sphere is $A = 4 \pi r^2. \\$Thus, the initial surface area $A_1 = 4 \pi (0.12)^2. \\$Using the Stefan-Boltzmann law for the initial conditions:$\\ P_1 = \sigma A_1 T_1^4 \\ 450 = \sigma \cdot 4 \pi (0.12)^2 \cdot (500)^4 \\$Now, consider the new conditions:$\\$• New radius $r_2 = \frac{r_1}{2} = 0.06$ m$\\$• New temperature $T_2 = 2 \times T_1 = 1000$ K$\\$The new surface area $A_2 = 4 \pi (0.06)^2. \\$Using the Stefan-Boltzmann law for the new conditions:$\\ P_2 = \sigma A_2 T_2^4 \\ P_2 = \sigma \cdot 4 \pi (0.06)^2 \cdot (1000)^4 \\$We can express $P_2$ in terms of $P_1$ by taking the ratio:$\\ \frac{P_2}{P_1} = \frac{\sigma \cdot 4 \pi (0.06)^2 \cdot (1000)^4}{\sigma \cdot 4 \pi (0.12)^2 \cdot (500)^4} \\$Simplify the expression:$\\ \frac{P_2}{450} = \frac{(0.06)^2 \cdot (1000)^4}{(0.12)^2 \cdot (500)^4} \\ \frac{P_2}{450} = \frac{0.0036 \cdot 10^{12}}{0.0144 \cdot 6.25 \times 10^{10}} \\ \frac{P_2}{450} = \frac{0.0036 \cdot 160000}{0.0144} \\ \frac{P_2}{450} = \frac{576}{0.0144} \\ P_2 = 450 \times 4 = 1800 \\$Therefore, the power radiated when the radius is halved and the temperature is doubled is $1800$ W.$\\$This corresponds to Option 3.