- assume forward biased
- replace the diode with iD going down and short circuit it and make VD=0
- combine 70Ω and 30Ω resistors that are parallel and it becomes 21Ω
- now 21Ω resistance is in parallel with 3V and the (40Ω resistors and 50mA)
- ID=50mA meaning the assumption was correct
It would probably be better to start by assuming the diode is open. This isolates the current source and the two series resistors, making it easy to check whether the anode of the diode is positive or negative with respect to the cathode.
Note that the 30Ω resistor is completely irrelevant to the rest of the circuit, since it is connected directly across a voltage source. You can ignore it altogether.
Note also that the power in the 40Ω resistor is determined entirely by the current source; nothing else in the circuit matters.