When the current reaches 0.1 mA the voltage across D3 and D4 will be 1.4 V. Assuming ideal diodes the voltage across D3 and D4 will remain at 1.4 V above 0.1 mA of diode current. Where will the current reach 0.1 mA? The answer is, when (Va-1.4)/(5000 + 1000) = 0.1 mA. Solving you find that when Va reaches 2 V then the diodes will be fully forward biased. So between 0 and 1.4 V on the input (i.e. Va) we can assume little to no current flow meaning Vb = Va and Vc = 0. At about 1.4 Va to 2 Va the diodes will begin to conduct. Above Va = 2 V the diodes will be fully saturated and what you have here can be thought of as a voltage divider output. So at Va=2V Vb will be 1.5V and Vc will be at 0.1 V. At any Va above 2 V the portion of the sine wave above 2V will be divided by (5000 + 1000)/1000 = 6. So any Va above 2 V will increase Vb above 2 V by an amount = (Va - 2 V)/6. Vc will increase from 0.1 V by an equal amount since the voltage across the diodes is assumed to be a constant 1.4 V.

During the negative portion of the sine wave D3 and D4 will be reversed biased and no current flows through the lower resistor. So again Vb = Va and Vc will be 0 V when Va is greater than -1.4 V. When Va goes more negative than -1.4 V D1 and D2 will begin to conduct. Here there is no voltage division so Vb will just be clipped and remain at about -1.4 V.

__Vb__:

To summarize, Vb will mimic Va between +1.4 Va and -1.4 Va. Between Va = 1.4 V and Va = 2.0 V, Vb will rise to 1.5 V. Above Va = 2 V, Vb will rise from 1.5 V to 1.5+(Va-2)/6. So for a 5 volt sine wave input the peak voltage will be reached at about 1.5 + (5-2)/6 = 2.0 VDC. Below -1.4 Va, Vb will remain at -1.4 VDC.

Vc:

From Va = 0 to Va = 1.4 V, Vc = 0 Volts. Between Va = 1.4 V and Va = 2.0 V, Vc will rise to 0.1 V. Above Va = 2.0 V, Vc will rise from 0.1 V to 0.1+(Va-2)/6. So for a 5 Volt sine wave input the peak voltage will be reached at about 0.1 + (5-2)/6 = 0.6 V. Below -1.4 Va, Vc will also remain at 0 V.