Answer
$\text{The ratio of the diameters is the same as the ratio of radii, so}$
$$\frac{D_{1}}{D_{2}}=[0.13]$$
Work Step by Step
$\text {From Wien's law (Eq. } 27-2) \text{we know that }$ $\lambda_{\mathrm{P}} T=\alpha,$ where $\alpha$ is a constant, so $$\lambda_{\mathrm{Pl}} T_{1}=\lambda_{\mathrm{P} 2} T_{2} .$$
$\text {From The Stefan- Boltzmann equation (Eq. } 14-6)$ we know that the power output of a star is given by $$P=\beta A T^{4},$$ where $\beta$ is a constant and $A$ is the radiating area.
the following informations are given:
$$b_{1} / b_{2}=0.091, d_{1}=d_{2}, \lambda_{\mathrm{Pl}}=470 \mathrm{nm},\quad and \quad \lambda_{\mathrm{P} 2}=720 \mathrm{nm} .$$
$$\lambda_{\mathrm{P} 1} T_{1}=\lambda_{\mathrm{P} 2} T_{2} \quad \rightarrow \quad \frac{T_{2}}{T_{1}}=\frac{\lambda_{\mathrm{Pl}}}{\lambda_{\mathrm{P} 2}} ; $$$$\quad b_{1}=0.091 b_{2} \quad \rightarrow \quad \frac{L_{1}}{4 \pi d_{1}^{2}}=0.091 \frac{L_{2}}{4 \pi d_{2}^{2}} \rightarrow$$
$$1=\frac{d_{2}^{2}}{d_{1}^{2}}=\frac{0.091 L_{2}}{L_{1}}=\frac{0.091 P_{2}}{P_{1}}=\frac{0.091 A_{2} T_{2}^{4}}{A_{1} T_{1}^{4}}=$$$$\frac{(0.091) 4 \pi r_{2}^{2} T_{2}^{4}}{4 \pi r_{1}^{2} T_{1}^{4}}=0.091 \frac{T_{2}^{4}}{T_{1}^{4}} \frac{r_{2}^{2}}{r_{1}^{2}} \rightarrow$$
$$\frac{r_{1}}{r_{2}}=\sqrt{0.091}\left(\frac{T_{2}}{T_{1}}\right)^{2}=\sqrt{0.091}\left(\frac{\lambda_{\mathrm{P} 1}}{\lambda_{\mathrm{P} 2}}\right)^{2}=$$$$\sqrt{0.091}\left(\frac{470 \mathrm{nm}}{720 \mathrm{nm}}\right)^{2}=0.1285$$
$\text{The ratio of the diameters is the same as the ratio of radii, so}$
$$\frac{D_{1}}{D_{2}}=[0.13]$$
