The RMS value of conduction current in a parallel plate capacitor is 6 . 9   μ A . The capacity of…

$I_{\mathrm{rms}}=6.9 \mathrm{\mu A}; V_{\mathrm{rms}}=230 \mathrm{V}; \omega =600 \mathrm{rad} {\mathrm{s}}^{-1}$

Current in capacitor $I_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{X_C}$, where, $X_C$ is capacitive reactance given by $X_C=\frac{1}{\omega C}$

Now, current, $I_{\mathrm{rms}}=\left(V_{\mathrm{rms}}\right)\times \left(\omega C\right)$

So, capacitance is $C=\frac{I_{\mathrm{rms}}}{V_{\mathrm{rms}}\omega }=\frac{6.9\times {10}^{-6}}{230\times 600}=50 \mathrm{pF}$