we know the energy of electron in n th orbit of Hydrogen
En = -K/n^2ev, k = constant
So the difference in the energy of the electron in n=2 and n=3 orbits is
E = E3 – E2 = -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