Power that is consumed by diode?
I think that you have neglected the voltage drop across the diode which limits the current. 

My solution is :
$$
\because V_{R_1}=V_0-V_f=5-2=3\,\text{V}
$$
$$
\therefore I_f=I_{R_1}=V_{R_1}/R_1=3/1000=3\,\text{mA}
$$
$$Power Loss =V_f\times I_f=2 \times0.003=6\,\text{mW} $$