If you assume the diode is Silicon, you can assume about 0.6 volts will appear across it for smallish currents.
This changes, but it is near enough for this analysis.
OK say there is 0.6 volts across the diode, then there must be 9.4 volts across the resistor.
(10 volts minus 0.6 volts = 9.4 volts)
Now a 2.7 K resistor with 9.4 volts across it must be carrying a current of about 3.48 mA. 9 V / 2700 ohms = 3.48 mA.
Notice that even though we assumed 0.6 volts for the diode voltage, it would not have made much difference if we had assumed 0.5 volts or 0.7 volts. 9.5/2700 = 3.5 mA. 9.3/2700 = 3.44 mA