This is a deceptive question... the #"H"# atom has an external #2s# orbital, the #"Li"# atom has a #2s# valence orbital, and the #"Na"# atom has a #2s# core orbital, RELATIVE to its own energy levels.
Sure, #"H"# has a #2s# orbital state it has not accessed and so it has the highest #2s# energy level from its perspective... but it is not possible to know the actual value of the energy level without proper computations to obtain the orbital energies.
(For all we know, hydrogen atom could be small enough that an "external #2s# orbital" really isn't that high in energy.)
Instead, consider the #2s# orbital of ONLY #"H"# and #"Li"#.
#E_n("1e"^(-)"-""atom") = -"13.605 eV" cdot Z^2/n^2#
with #Z# being the atomic number and #n# the energy level for one-electron atoms ONLY.
For hydrogen atom this is easy:
#E_(2s)("H") = -"13.605 eV" cdot 1^2/2^2 = ul(-"3.401 eV")#
For lithium atom we have to be creative. Its ionization energy is #"5.392 eV"#, so its energy level is
#E_(2s)("Li") = ul(-"5.392 eV")#
by Koopman's approximation theorem for atoms with 1 valence electron.
We see that for the #2s# orbital, the lithium atom has a more core-like #2s# orbital, as its energy is more negative (actually, it's valence for lithium and external for hydrogen). That tells us that the energy levels converge (and become more numerous) as the atom size increases.
(By inference, sodium has a #2s# orbital energy that is even more negative.)
Therefore, #"H"# atom has the highest energy for its #2s# orbital by analogy.
