The total current in the circuit is\\\\(I = \\dfrac{2E}{r_1 + r_2 + R}\\\\) ..............................(1).....\\\\((\\\\)In series, \\\\(E + E = 2E)\\\\)Now the potential drop across first cell is\\\\(V_1 = E - Ir_1\\\\)As per the given condition\\\\(V_1 = 0\\\\)hence,\\\\(E - Ir_1 = 0\\\\)\\\\(E - \\left(\\dfrac{2E}{r_1 + r_2 + R} \\right)r_1 = 0\\\\) .....................from\\\\((1)\\\\)\\\\(\\dfrac{2r_1E}{r_1 + r_2 + R} = E\\\\)\\\\(\\Rightarrow r_1 + r_2 + R = 2r_1\\\\)\\\\(R = r_1 - r_2\\\\)