I will choose the bottom node as the reference node and assign node voltages V_A, V_B, V_C, and V_D to the remaining nodes.

Define vD (used to define the dependent current source) in terms of node voltages.

vD = V_C - V_B

Define if (used to define the dependent voltage source) in terms of node voltages.

if = V_C / 40

Node B forms a supernode with the reference node due to the 20 V source (shown with a dotted outline):

V_B = 20

Nodes A and D also form a supernode due to the dependent voltage source (shown with a dotted outline):

V_A - V_D = 35 if = 35 V_C / 40

40 V_A - 35 V_C - 40 V_D = 0 (Eq. 1)

KCL at supernode A-D:

Simplify and collect terms (LCD = 80)

- 4 V_A + 40 V_B - 40 V_A + 20 V_C - 20 V_D - V_D -250 VD = 0

- 4 V_A + 40 (20) - 40 V_A + 20 V_C - 20 V_D - V_D -250 (V_C - 20) = 0

- 44 V_A - 230 V_C - 21 V_D = - 5800 (Eq. 2)

KCL at node C:

Simplify and collect terms (LCD = 40)

40 V_B - 40 V_C + 10 V_D - 10 V_C - V_C = 0

40 (20) - 40 V_C + 10 V_D - 10 V_C - V_C = 0

- 51 V_C + 10 V_D = - 800 (Eq. 3)

Solving Equations 1 and 2 yields:

V_A = 29.54 V

V_B = 20 V

V_C = 18.33 V

V_D = 13.5 V

To find the power developed by the 20 V source, find the current leaving the + reference of that source. By KCL, this current is equal to the sum of the currents leaving node B through the 2 W and 1 W resistors. Thus the current up through the 20 V source is

I_s = 10 - 14.77 + 20 - 18.33 = - 3.1 A

So power developed is

P = 20 I_s = 20 (-3.1) = - 62 W

Thus the 20 V source is actually absorbing power.