${{V}_{3\Omega }}={{V}_{6\Omega }}\Rightarrow 3\times 2=6\times {I}'\Rightarrow {I}'=1\Rightarrow I=1+2=3A$ $\varepsilon ={{I}_{r}}={{V}_{2\Omega }}+{{V}_{3\Omega }}\Rightarrow \varepsilon -3\times 1=6+6\Rightarrow \varepsilon =15V$