Number of atoms in 2 kg deuterium = \\\\(\\dfrac{6.023\ imes 10^{23}\ imes 2000}{2}=6.023\ imes 10^{26}\\\\)Energy released when 2 atoms fuse = \\\\(3.27MeV\\\\)Thus, total energy released = \\\\(\\dfrac{3.27}{2}\ imes 6.023\ imes 10^{26}MeV\\\\)\\\\(=15.75\ imes 10^{13}J\\\\).Energy consumed by bulb per second=\\\\(100J\\\\).Thus, time for which the bulb glows = \\\\(\\dfrac{15.75\ imes 10^{13}}{100}s\\\\)\\\\(=15.75\ imes 10^{11}s=4.99\ imes 10^7 years\\\\).