The given circuit can be drawn as
Applying Kirchhoff's voltage law (KVL) in loop ABCDA , we get −(i1+i2)R+E2−i2r2=0...(i)
Applying KVL in loop ABFEA , we get −(i1+i2)R−i1r1+E1=0...(ii)
Applying KVL in loop EFCDE, we get −E1+i1r1+E2−i2r2=0...(iii)
From the given options, only option (c) satisfies Eq. (ii), hence it is the correct equation.