First test the diode-off state. The resistors then form a 2 kΩ+6 kΩ series circuit across 2 V. Its current is $2/8000=0.25$ mA.
The voltage across the 2 kΩ resistor, and hence the diode, is $0.25\cdot2=0.5$ V. This is below the 0.7 V turn-on voltage, so the off assumption is consistent. Thus **$i_2=0.25$ mA**.
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