In the given circuit diagram, a wire is joining points B and D. The current in this wire is:

Taking node C as reference: $$V_C = 0 \implies V_A = 20$$

Since points B and D are connected by a wire: $$V_B = V_D = V$$

Kirchhoff's current law at node B + D combined:

$$\frac{V - 20}{1} + \frac{V - 20}{4} + \frac{V - 0}{2} + \frac{V - 0}{3} = 0$$

$$V\left(1 + \frac{1}{4} + \frac{1}{2} + \frac{1}{3}\right) = 20\left(1 + \frac{1}{4}\right)$$

$$V\left(\frac{12 + 3 + 6 + 4}{12}\right) = 20\left(\frac{5}{4}\right) \implies V\left(\frac{25}{12}\right) = 25 \implies V = 12$$

Currents entering node B from node A and leaving to node C: $$I_{AB} = \frac{20 - 12}{1} = 8$$

$$I_{BC} = \frac{12 - 0}{2} = 6$$

Kirchhoff's current law at node B: $$I_{AB} = I_{BC} + I_{BD} \implies 8 = 6 + I_{BD} \implies I_{BD} = 2$$