we know the energy of electron in n th orbit of Hydrogen

E_{n} = -K/n^2ev, k = constant

So the difference in the energy of the electron in n=2 and n=3 orbits is

E = E_{3} – E_{2 }= -K[1/3^2-1/2^2] = K*5/36

=>K = 7.2E

Now ionization energy of hydrogen atom is the energy required to promote electron from n=1 to n= ∞

So This ionization energy = -K{1/∞^2-1/1^2]=K =7,2E

Hence option (B) is accepted