Nodal analysis current in 6 Ω

Circuit reading (from the figure). Node C is joined to the bottom rail, so take C as the reference (0 V). The 2 A source and the 12 Ω resistor $R_1$ are both between node D and the rail. Between D and B there are two paths in parallel: the 6 Ω bridge $R_4$ and the series path D–A–B ($8+4=12\ \Omega$; node A has no other connection, so no separate equation is needed). The 20 Ω resistor joins D to the reference and the 10 Ω resistor joins B to the reference.

Unknowns: $V_D$ and $V_B$.

KCL at node D (2 A flows in, the rest leaves through the branches):

$$2=\frac{V_D}{12}+\frac{V_D}{20}+\frac{V_D-V_B}{6}+\frac{V_D-V_B}{12}$$

KCL at node B:

$$\frac{V_B-V_D}{6}+\frac{V_B-V_D}{12}+\frac{V_B}{10}=0$$

Collect terms (conductances in siemens):

$$\begin{aligned} 0.3833\,V_D-0.25\,V_B&=2\\ -0.25\,V_D+0.35\,V_B&=0 \end{aligned}$$

From the second equation, $V_B=0.7143\,V_D$. Substituting in the first:

$$0.3833V_D-0.1786V_D=2 \Rightarrow V_D=9.767\ \text{V},\quad V_B=6.977\ \text{V}$$

Current through the 6 Ω resistor:

$$I_{6\Omega}=\frac{V_D-V_B}{6}=\frac{9.767-6.977}{6}=0.465\ \text{A}\ (D\to B)$$

Answer: $I_{6\Omega}=0.465$ A, from D to B.