Beginner Thevenin equivalent diagram
I can't really reproduce what you were calculating, maybe you can explain. The steps you need to take to get the current through the 2\$\Omega\$ resistor


Calculate the voltate drop across the 4\$\Omega\$ by merging \$(2+5)||4 = 2.55\Omega\$ which then form a voltage divider with the other two resistors and gives a voltage across the 4\$\Omega\$ of \$9V/12.55*2.55=1.82V\$
Knowing the voltage across the 5\$\Omega\$ + 2\$\Omega\$ you can calculate the current \$I_{2\Omega}\$ as \$1.82V/7\Omega=261.24mA\$ (without the rounding you get the 260.8mA from your solution


The bad thing is that this has nothing to do with Thevenin.
To solve it using Thevenin you would remove your load resistor from the circuit







simulate this circuit &ndash; Schematic created using CircuitLab

You can now calculate the output voltage or \$V_{th}\$ as \$9V/14\Omega*4\Omega=2.571V\$. 

Then short circuit your voltage supply to calculate \$R_{th}\$ as
$$
((4\Omega+6\Omega) || 4\Omega) + 5\Omega = 7.8571\Omega
$$

and finally your load current is \$V_{th}/(R_{th}+R_{load}=260.8mA\$