Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We are given a simple $RC$ circuit at which the capacitor starts to discharge when the switch is closed at $t=0$ s where the decaying current in this circuit is given by
$$I=\dfrac{(\Delta V_C)_{\rm max}}{R}e^{-t/RC}\tag 1$$
As we know, the magnetic field inside a capacitor is given by
$$B=\dfrac{\mu_0}{2\pi}\dfrac{r}{R^2}\dfrac{dQ}{dt} $$
See Figure 34.18 in your textbook.
Recalling that $dQ/dt=I$,
$$B=\dfrac{\mu_0}{2\pi}\dfrac{r}{R^2} I$$
Plug $I$ from (1),
$$B=\dfrac{\mu_0}{2\pi}\dfrac{r}{R_{\rm plate}^2} \dfrac{(\Delta V_C)_{\rm max}}{R}e^{-t/RC} \tag 2$$
Finding the magnitude of $C$; where the capacitance is given by $C=\epsilon_0 A/d$ where $A=\pi R^2$ where $R$ is the radius of the capacitor's plates.
$$C=\dfrac{\epsilon_0 \pi R^2}{d}=\dfrac{(8.85\times 10^{-12}) \pi (0.02)^2}{(0.001)}=\bf 1.11212\times 10^{-11}\;\rm F$$
Plug the known into (2);
$$B=\dfrac{(4\pi\times 10^{-7})}{2\pi }\dfrac{(0.01)}{(0.02)^2} \dfrac{(1000) }{ (0.2)}e^{-t/(0.2)(1.11212\times 10^{-11})}$$
$$\boxed{B=0.025e^{-t /(2.224\times 10^{-12})}}$$
$$\color{blue}{\bf [b]}$$
See the graph below,
