The plates of a capacitor are discs of radius 8.0 cm, with a vacuum between them…

Step 1. The displacement current. $I_d = \varepsilon_0 \pi R^2\dfrac{dE}{dt} = 8.85 \times 10^{-12} \times \pi \times (0.080\ \text{m})^2 \times 2.0 \times 10^{12}\ \text{V/(m s)} = 0.356\ \text{A}$.

Step 2. The edge. $B = \dfrac{\mu_0 I_d}{2\pi R} = \dfrac{2 \times 10^{-7}\ \text{T m/A} \times 0.356\ \text{A}}{0.080\ \text{m}} = 0.890\ \mu\text{T}$.

Step 3. The circle of radius $4.0$ cm. The fraction is $\left(\dfrac{4.0}{8.0}\right)^2 = 0.25$, so $I_{\text{in}} = 0.0890$ A. Then $B = \dfrac{2 \times 10^{-7}\ \text{T m/A} \times 0.0890\ \text{A}}{0.040\ \text{m}} = 0.445\ \mu\text{T}$, half of the field at the edge.