Superposition principle and voltage division
Hmm. 

First I remove \$2V_a\$ source. So we left with this:

 

And we can find \$V_{O1}\$

$$V_{O1} = V_{SS} \frac{2R_a}{R+ 2R_a} = 0.081V = 81mV$$

Nowe we turn-off the \$V_{SS}\$ source and we are ending with this circuit: 



And now we can solve for \$V_{O2}\$ 

As you can see this time \$V_{O2}\ = I\cdot R\$  or  \$V_{O2} = 2V_a - I\cdot 2R_a\$

Where:

\$I = \frac{2V_a}{R+2R_a}\$ therefore: 

$$V_{O2} =I\cdot R =\frac{2V_a}{R+2R_a}\cdot R = 2V_a \frac{R}{R+2R_a} = 1.167V $$

and finally 

$$V_O =V_{O1}+V_{O2} = V_{SS} \frac{2R_a}{R+ 2R_a} + 2V_a \frac{R}{R+2R_a} =1.248V   $$

Your circuit looks like this:



I mark the voltage drop by the arrows and the arrow tip is pointing the "positive" side.  

And from KVL we have: 

$$2V_a = V_{2R_a} + V_R$$  and also notice that \$V_o2\$ is a voltage drop betwenn \$V_A\$ node and GND.  

$$V_A = 2Va - V_{2R_a} = 2Va - I*{2R_a} $$ or becouse \$V_A = V_R = I*R\$ 

Do you see it?