The RMS value of conduction current in a parallel plate capacitor...

${I}_{rms} = 6 . 9\mu A; {V}_{rms} = 230V; \omega = 600rad{s}^{- 1}$

Current in capacitor ${I}_{rms} = \dfrac{{V}_{rms}}{{X}_{C}}$, where, ${X}_{C}$ is capacitive reactance given by ${X}_{C} = \dfrac{1}{\omegaC}$

Now, current, ${I}_{rms} = \left({V}_{rms}\right) \times \left(\omegaC\right)$

So, capacitance is $C = \dfrac{{I}_{rms}}{{V}_{rms}\omega} = \dfrac{6 . 9 \times {10}^{- 6}}{230 \times 600} = 50pF$