Not sure where you made a mistake. It seems like you only have three terms in your sum, but there are four legs. Is that part of the problem? I actually think this is a hard problem, but I graduated from college over 10 years ago. Let me walk you through it.
I am going to sum all the currents going into V2. Just have to calculate them first, and write them with the correct polarity (going INTO V2).
i_1 is -2A by inspection.
So 2*(i_1) is -4A.
Current in the Voltage source is V1/100 +2A (going up... this itself is a sum)
Current in the 110 Ohm is (V1-20)/110 (going to right in 110 Ohm resistor)
So, summing currents into node, -2A -4A +2A + V1/100 + (V1-20)/110 = 0.
V1/100 + (V1-20)/110 = 4
V1/100 + V1/110 - 20/110 = 4
V1/100 + V1/110 = 4+20/110
V1 (1/100 + 1/110) = 4 + 20/110
V1 = (4 + 20/110)/(1/100 + 1/110)
V1 = 219.05.
Hope that helps.