(a)
Hint: \(I_{rms}=\frac{V}{X_{C}}\)
Step 1: Find the capacitive reactance.
\(X_{C}=\frac{1}{\omega C}=\frac{1}{300\times 100\times 10^{-12}}=\frac{1}{3}\times 10^{8}~\Omega\)
Step 2: Find RMS value of conduction current.
The radius of each circular plate, $\mathrm{R}=6 \mathrm{cm}=\frac{6}{100} \mathrm{m}=0.06\mathrm{m}$
RMS value of conduction current, ${\mathrm{I}}_{\mathrm{rms}}=\frac{\mathrm{V}}{{\mathrm{X}}_{\mathrm{c}}}$

$\Rightarrow Irms=V\times \omega C=230\times 3\times 10-8=6.9\times 10-6A=6.9\mu A$

(b)
Hint: Recall the relation between conduction and displacement current.
Yes, the conduction current is equal to displacement current.

(c)
Hint: Use the relation between magnetic field and peak value of current.
Step 1: Find peak value of current.
\(I_{0}=\sqrt{2}I_{rms}=\sqrt{2}\times6.9\times10^{-6}~A\)

Step 2: Find magnetic field.
Magnetic field is given by: $\mathrm{B}=\frac{{\mathrm{\mu }}_{\circ }\mathrm{r}}{2{\mathrm{\pi R}}^2}{\mathrm{I}}_0=\frac{{\mathrm{\mu }}_{\circ }\mathrm{r}}{2{\mathrm{\pi R}}^2}\sqrt{2}{\mathrm{I}}_{\mathrm{rms}}$
where r = Distance between the plates from the axis = 3.0 cm = 0.03 m

$$B$=\frac{4\pi \times {10}^{-7}\times 0.03\times \sqrt{2}\times 6.9\times {10}^{-6}}{2\pi {\left(0.06\right)}^2}=1.63\times {10}^{-11}T$