At node A

$\dfrac{V_A-6}{3}+\dfrac{V_A}{6}+\dfrac{V_A-V_B}{4}=0 \\ V_A[3^{-1}+6^{-1}+4^{-1}]-V_B(4^{-1})=2 \\ 0.75V_A-0.25V_B=2....................(1)$

At node B

$\dfrac{V_BV_A}{4}+\dfrac{V_B}{2}-5=0 \\ -(4^{-1})V_A+(4^{-1}+2^{-1})V_B=5 \\ -0.25V_A+0.75V_B=5.................(2)$

From (1) & (2)

$VA=5.5 V \\ VB=8.5 V \\ I_{4\Omega}=\dfrac{V_A-V_B}{4}=-0.75 \\ I_{4\Omega}=0.75(\leftarrow)$