A capacitor of 12 µF is charged so that the voltage across it rises steadily at 25 V/s…

Step 1. The charging current. $I = C\dfrac{dV}{dt} = 12 \times 10^{-6}\ \text{F} \times 25\ \text{V/s} = 0.30 \times 10^{-3}\ \text{A} = 0.30\ \text{mA}$.

Step 2. The displacement current. It equals the charging current: $I_d = 0.30$ mA. So the total current has no break between the wire and the gap.

Step 3. The rate of change of the flux. We want $d\Phi_E/dt$, so $\dfrac{d\Phi_E}{dt} = \dfrac{I_d}{\varepsilon_0} = \dfrac{0.30 \times 10^{-3}\ \text{A}}{8.85 \times 10^{-12}\ \text{C}^2/(\text{N m}^2)} = 3.39 \times 10^{7}\ \text{V m/s}$.