The user wants me to solve a physics problem involving blackbody radiation from two spherical stars, A and B. I need to find the ratio of their peak emission wavelengths, $\frac{\lambda_A}{\lambda_B}$.

Step-by-Step Solution:

1. Identify the relevant physical laws. The power radiated by a blackbody is described by the Stefan-Boltzmann Law, and the relationship between the peak wavelength of emission and temperature is given by Wien's Displacement Law.

2. Apply the Stefan-Boltzmann Law. The power (P) radiated by a spherical blackbody is given by: $P = \sigma A T^4 = \sigma (4\pi R^2) T^4$ where $\sigma$ is the Stefan-Boltzmann constant, $R$ is the radius, and $T$ is the surface temperature.

   For star A: $P_A = \sigma (4\pi R_A^2) T_A^4$ For star B: $P_B = \sigma (4\pi R_B^2) T_B^4$

3. Use the given information to find the ratio of temperatures. We are given:

   • The ratio of radii: $\frac{R_A}{R_B} = 400$
   • The ratio of emitted powers: $\frac{P_A}{P_B} = 10^4$

   Let's take the ratio of the powers of the two stars: $\frac{P_A}{P_B} = \frac{\sigma (4\pi R_A^2) T_A^4}{\sigma (4\pi R_B^2) T_B^4} = \left( \frac{R_A}{R_B} \right)^2 \left( \frac{T_A}{T_B} \right)^4$

   Substitute the given values into this equation: $10^4 = (400)^2 \left( \frac{T_A}{T_B} \right)^4$

   Now, solve for the ratio of temperatures, $\frac{T_A}{T_B}$: $\left( \frac{T_A}{T_B} \right)^4 = \frac{10^4}{(400)^2} = \frac{10^4}{(4 \times 10^2)^2} = \frac{10^4}{16 \times 10^4} = \frac{1}{16}$

   Taking the fourth root of both sides: $\frac{T_A}{T_B} = \left( \frac{1}{16} \right)^{1/4} = \frac{1}{2}$

4. Apply Wien's Displacement Law. Wien's law states that the product of the peak wavelength ($\lambda$) and the temperature (T) is a constant (b): $\lambda T = b$ This implies that the peak wavelength is inversely proportional to the temperature: $\lambda \propto \frac{1}{T}$.

   For star A: $\lambda_A T_A = b$ For star B: $\lambda_B T_B = b$

   Therefore, $\lambda_A T_A = \lambda_B T_B$.

5. Calculate the ratio of the wavelengths. Rearranging the equation from Wien's law: $\frac{\lambda_A}{\lambda_B} = \frac{T_B}{T_A}$

   We found that $\frac{T_A}{T_B} = \frac{1}{2}$. The reciprocal of this is: $\frac{T_B}{T_A} = 2$

   Therefore, the ratio of their wavelengths is: $\frac{\lambda_A}{\lambda_B} = 2$

Final Answer:

The ratio of their wavelengths $\left( {{{{\lambda _A}} \over {{\lambda _B}}}} \right)$ is 2.