The ripple voltage in a full wave rectifier can be calculated using the formula:

Vr = Vm / (2 * f * C * RL)

Where:
Vr is the ripple voltage
Vm is the peak voltage of the AC input signal
f is the frequency of the AC input signal
C is the capacitance of the smoothing capacitor
RL is the load resistance

In this question, the AC voltage applied to the diodes is given as 220V (rms), which means the peak voltage (Vm) is √2 times the rms voltage. So, Vm = 220V * √2 = 311.13V.

The frequency of the AC voltage is not given, so we'll assume a standard frequency of 50Hz.

The load resistance (RL) is given as 1K (1000 ohms).

Now, let's calculate the ripple voltage using the given values:

Vr = 311.13V / (2 * 50Hz * C * 1000 ohms)

Since the diode resistance is neglected, the load resistance (RL) is the only resistance in the circuit.

To simplify the calculation, we can assume a typical value for the smoothing capacitor (C) in a full wave rectifier circuit, which is usually in the range of 100-1000µF. Let's assume a value of 100µF (0.0001F).

Plugging in the values:

Vr = 311.13V / (2 * 50Hz * 0.0001F * 1000 ohms)

Vr = 311.13V / (0.01F * 1000 ohms)

Vr = 311.13V / 10 ohms

Vr = 31.113V

Therefore, the ripple voltage is approximately 31.113V.

However, none of the given options match this value. So, it seems there may be a mistake in the question or options provided.

If we assume that the correct value of the ripple voltage is 0.954V (option C), then we can calculate the required capacitance (C) using the formula as follows:

0.954V = 311.13V / (2 * 50Hz * C * 1000 ohms)

Solving for C:

C = 311.13V / (2 * 50Hz * 0.954V * 1000 ohms)

C = 311.13V / (2 * 50Hz * 0.954V * 1000 ohms)

C = 311.13V / (0.0954F * 1000 ohms)

C = 311.13V / 95.4 ohms

C ≈ 3.26µF

So, if the ripple voltage is indeed 0.954V, then the required capacitance to achieve this ripple voltage is approximately 3.26µF.