Thévenin's Theorem Example
Since you are just learning about Thevenin, it's probably easiest to see things in the following way:


Notice that the \$6\:\text{k}\Omega\$ resistor and the \$4\:\text{k}\Omega\$ resistor span between two voltage sources (at \$0\:\text{V}\$ and \$20\:\text{V}\$.)
Convert that pair of resistors and voltage sources to their Thevenin equivalent (which will be a new voltage source and a series resistance.)
Now you should have a Thevenin voltage, \$V_\text{TH}\$, followed by its Thevenin resistance, \$R_\text{TH}\$, followed by a \$4.9\:\text{k}\Omega\$ resistor (in series), followed by the diode.
Since you know the diode drop (given to you, a priori), you just subtract it from the Thevenin voltage value. This will be the remaining voltage that is across the remaining Thevenin resistance in series with the \$4.9\:\text{k}\Omega\$ resistor.
The problem is now reduced to \$I_D=\frac{V_\text{TH}-V_D}{R_\text{TH}+4.9\:\text{k}\Omega}\$.


That's all there is to it.

The primary insight is recognizing step (1) above. Those two resistors and their voltage sources can be converted readily into \$V_\text{TH}\$ and \$R_\text{TH}\$. The rest is just basic machinery steps.