According to the given figure
Consider $I_d$ be the displacement current in the region between two plates of parallel plate capacitor, shown in figure.
The magnetic field induction at a point in a region between two plates of capacitor at a perpendicular distance $r$ from the axis of plates is
$$
\begin{array}{rlr}
\mathrm{B} & =\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{r}}=\frac{\mu_0}{2 \pi \mathrm{r}} \mathrm{I}_{\mathrm{d}}=\frac{\mu_0}{2 \pi \mathrm{r}} \times\left(\varepsilon_0 \frac{\mathrm{d} \phi_{\mathrm{E}}}{\mathrm{dt}}\right) & {\left[\because \mathrm{I}_{\mathrm{d}}=\frac{\mathrm{E}_0 \mathrm{~d} \phi_{\mathrm{E}}}{\mathrm{dt}}\right]} \\
= & \frac{\mu_0 \varepsilon_0}{2 \pi \mathrm{r}}\left[\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{E} \pi \mathrm{r}^2\right)\right]=\frac{\mu_0 \varepsilon_0}{2 \pi \mathrm{r}} \pi \mathrm{r}^2 \frac{\mathrm{dE}}{\mathrm{dt}} & \\
\mathrm{B}=\frac{\mu_0 \varepsilon_0 \mathrm{r}}{2} \frac{\mathrm{dE}}{\mathrm{dt}} & {\left[\because \phi_{\mathrm{E}}=\mathrm{E} \pi \mathrm{r}^2\right]}
\end{array}
$$
Hence Proved.