node analysis with current source
But you first wrote I2 = I1 + I3 haven't you ??. All this means that you have assumed that I1 and I2 current flow into VB node. 
I3 is flowing into VB node. And this current is equal to 0.1 A (current source current) because in series circuit we have the same current everywhere and this is why R3 and R4 do not have any influence over I3 current, only current source count here. I1 also flows into VB node and this means that the I1 current is flowing from positive side of the Voltage source through R1 into VB node.
And this is why we have I1 = (E - VB)/R1. 

$$I2= I1 + I3$$
$$\frac{VB}{R2}= \frac{E- VB}{R1} + I $$
$$\frac{VB}{600\Omega}= \frac{48V- VB}{400\Omega} + 0.1A $$

Of course, you could write it this way also (all current flow out of the VB node, and that means that I assumed that VB node has the highest voltage/electric potential) 

$$I1+I2+I3=0$$ 

$$\frac{VB - E}{R1}+\frac{VB}{R2}+(-I)=0$$ 

$$\frac{VB - 48V}{400\Omega}+\frac{VB}{600\Omega}-0.1A=0$$ 

and the solution is $$VB = 52.8V$$