In a round region of radius 20 cm the electric field is uniform and changes as E = E₀sinω t, with…

Step 1. The inner circle, $r = 0.10$ m. The flux is $E\pi r^2$, so $B\cdot 2\pi r = \dfrac{1}{c^2}\pi r^2\,\omega E_0$. Then $B = \dfrac{r\omega E_0}{2c^2}$. The top is $r\omega E_0 = 1.508 \times 10^{11}$ and the bottom is $2c^2 = 1.8 \times 10^{17}$, in SI units. So $B = 0.838\ \mu\text{T}$.

Step 2. The outer circle, $r = 0.30$ m. The field exists only in the region of radius $R$, so the flux is $E\pi R^2$. Then $B = \dfrac{R^2\omega E_0}{2rc^2}$. The top is $R^2\omega E_0 = 6.03 \times 10^{10}$ and the bottom is $2rc^2 = 5.4 \times 10^{16}$, in SI units. So $B = 1.117\ \mu\text{T}$.

Step 3. The ratio. $\dfrac{1.117}{0.838} = 1.33$.