Three identical p-n junction diodes D_1 , D_2 and D_3 are connected...

From the given circuit diagram, we can determine the biasing of each diode by observing their connections to the battery.

The battery symbol has its longer line on the left and shorter, thicker line on the right. This indicates that the left side of the circuit is connected to the positive terminal and the right side to the negative terminal.

1. For diode $D_1$: The p-side (triangle) is connected to the positive terminal (via a resistor) and the n-side (straight line) is connected to the negative terminal. Thus, $D_1$ is forward-biased.
2. For diode $D_2$: The p-side is connected to the positive terminal, but there is a clear break in the circuit on its n-side. Since it is an open circuit, no current flows through it, meaning $D_2$ is unbiased.
3. For diode $D_3$: The n-side is connected to the positive terminal and the p-side is connected to the negative terminal (via a resistor). Thus, $D_3$ is reverse-biased.

The width of the depletion region of a p-n junction depends on its biasing:
- In forward bias, the applied voltage opposes the built-in potential, causing the depletion width to decrease.
- In an unbiased state, the depletion width remains at its natural equilibrium value.
- In reverse bias, the applied voltage adds to the built-in potential, causing the depletion width to increase.

Let $W_{eq}$ be the equilibrium depletion width. Based on the biasing, we have:
$W_1 < W_{eq}$ (forward-biased)
$W_2 = W_{eq}$ (unbiased)
$W_3 > W_{eq}$ (reverse-biased)

Comparing these widths, we get the relation:
$W_3 > W_2 > W_1$

Answer: $W_3>W_2>W_1$