In the circuit shown in figure, $E_{1}and{E}_2$ are two ideal sources of unknown emfs. Some currents are shown. Potential difference appearing across  $6\Omega$ resistance is $V_A-V_B=10V.$ 

After redrawing the circuit

$\left(1\right)I_4=5A$

$\left(2\right)Fromloop\left(1\right),$

$-8\left(3\right)+E_1-4\left(3\right)=0or{E}_1=36V$

From loop (2), 

$+4\left(5\right)+5\left(2\right)-E_2+8\left(3\right)=0$

$or{E}_2=54V$

$\left(3\right)Fromloop\left(3\right),$

$-2R-E_1+E_2=0$

$orR=\frac{E_2-E_1}{2}=-36=9\Omega$