I will choose node d as the reference node and assign node voltages V_a, V_b, and V_c to nodes a, b, and c respectively. Thus the equations are:

KCL at node A:

First, determine V_2a so that we can get an expression for the current into node A through the 2 W resistor.

V_B - V_A = - 110 + V_2a

V_2a = V_B - V_A + 110

Now write KCL into node A

Simplify and collect terms.

8 V_B - 8 V_A + 880 + 2 V_C - 2 V_A - V_A = 0

- 11 V_A + 8 V_B + 2 V_C = - 880 (Eq. A)

KCL at node B:

The current into node B through the upper 110 V source is simply the negative of the current through the 2 W resistor found above, so I'll skip the gory details. The current into node B through the lower 110 V source requires that we find V_2b, however.

- V_B = V_2a - 110

V_2a = 110 - V_B

Now KCL into node B

Simplify and collect terms

3 V_A - 3 V_B - 330 + 330 - 3 V_B + 2 V_C - 2 V_B = 0

3 V_A - 8 V_B + 2 V_C = 0 (Eq. B)

KCL at node C:

Simplify and collect terms.

3 V_A - 3 V_C + 8 V_B - 8 V_C - V_C = 0

3 V_A + 8 V_B - 12 V_C = 0 (Eq. C)

Solving Equations A, B, and C yields:

V_A = 157.1 V

V_B = 82.5 V

V_C = 94.29 V