Q:  What is the ratio of the number of neutral atoms in the 
           (3p) excited state to those in the (3s) ground state?


          Nb       6    - ( 3.37 x 10^(-19) J /  8.00 x 10^(-20) J )
         ----  =  ---  e
          Na       2

                         -4.21
               =   3   e  


               =   0.05  


        That's ... a small number.