Solution:

Radius of each circular plate is given as $0.06m$

Capacitance of a parallel plate capacitor is given as $\mathrm{C}=100 \mathrm{pF}=100 \times 10^{-12} \mathrm{~F}$

Supply voltage is given as $V=230 \mathrm{~V}$

Angular frequency is given as $\omega=300 \mathrm{rad} \mathrm{s}^{-1}$

Magnetic field is represented by the equation,

$B=\frac{\mu_{0} r}{2 \pi R^{2}} I_{0}$

Where,

$\mu_{0}$ is the permeability of free space with a value of $4 \pi \times 10^{-7} N A^{-2}$

$I_{0}$ is the maximum value of current with a value of $\sqrt{2} I$

$r$ is the distance between the plates from the axis with a value of $0.03m$

$\therefore B=\frac{4 \pi \times 10^{-7} \times 0.03 \times \sqrt{2} \times 6.9 \times 10^{-6}}{2 \pi \times(0.06)^{2}}$

$=1.63 \times 10^{-11} \mathrm{~T}$

As a result, $1.63 \times$$10^{-11} T$ is the the magnetic field at that point.