Given - initial voltage  $$V_1 = V,$$ Initial charge $$q_1 = 360 mu C$$​ Therefore , capacitance $$C = dfrac{q_1}{V_1} =dfrac{360}{V}$$ -- (i) When , voltage  $$V_2 = V - 120,$$ charge $$q_2 = 120 mu C$$  Therefore , capacitance $$C = dfrac{q_2}{V_2} = dfrac{120}{V-120}$$ --(ii) Capacitance of the capacitor remains constant as there is no change in area of the plates , medium and distance between plates , (i) Dividing (i) by (ii) , $$dfrac{360}{V}=dfrac{120}{(V−120)}$$ $$or, 3(V−120)=V$$ $$or, V=360/2=180volt$$ (ii)Capacitance  $$C=360/180=2μF$$