(I) Consider 4A current source.

$I_{6\Omega}^1=\dfrac{I_1(5)}{6+5}.......(1)$

$(5^{-1}+6^{-1})^{-1}=2.727 \Omega$

$(2+2.727)=4.727 \Omega$

$$I_1=\dfrac{4(10)}{[10+4.727]}=2.716 A \\ From(1) \\ I_{6\Omega}^1=\dfrac{(2.716)(5)}{11}1.234 A (\downarrow)$$

II) Consider 10V voltage source:

$(5^{-1}+6^{-1})^{-1}=2.727 \Omega$

$$I=\dfrac{10}{(10+4.727)}=0.679 A \\ I_{6\Omega}^{11}=\dfrac{I(5)}{(6+5)}=0.3086A(\uparrow)$$

III) Consider 3A current source;

$$I_{6\Omega}^{111}=\dfrac{3(3.529)}{6+3.259}=1.111A(\downarrow)$$

IV) By using superposition theorem

$I6Ω1 = (1.234-0.3086+1.111) A(↓) \\ I6Ω = 2.0364 A$