To solve this problem, we need to understand how to calculate the rms current in an AC circuit containing a capacitor, as well as the concept of displacement current.

First, the rms (root mean square) value of the current ($I_{\text{rms}}$) in a capacitor when connected to an AC supply is given by:

$I_{\text{rms}} = V_{\text{rms}} \cdot \omega C$

where $V_{\text{rms}}$ is the rms voltage, $\omega$ is the angular frequency, and $C$ is the capacitance.

For an AC supply, the rms voltage is related to the peak voltage ($V_0$) by:

$V_{\text{rms}} = \frac{V_0}{\sqrt{2}}$

However, we've been given the rms voltage directly, which is $230\, \text{V}$. Therefore, we can use this value directly in our calculations.

Given:

$C = 200\, \text{pF} = 200 \times 10^{-12}\, \text{F}$

$\omega = 300\, \text{rad/s}$

$V_{\text{rms}} = 230\, \text{V}$

Substitute these values into the equation for rms current:

$I_{\text{rms}} = V_{\text{rms}} \cdot \omega C = 230 \times 300 \times 200 \times 10^{-12} = 13.8 \times 10^{-6}\, \text{A}$

Therefore, the rms value of the conduction current is $13.8\, \mu \text{A}$.

Now, for the displacement current (which is essentially the current that would "flow" through the dielectric of the capacitor as a result of the changing electric field). In an AC circuit, the displacement current and the conduction current are the same. That's because the displacement current is necessary to sustain the changing electric field between the plates of the capacitor, and this changing field is what causes the conduction current in the leads and the rest of the circuit.

Hence, the rms value of the displacement current is the same as the conduction current, which is $13.8\, \mu \text{A}$.

So, the correct option is:

Option B

$13.8\, \mu \text{A}$ and $13.8\, \mu \text{A}$