Since Lithium 2+ (atomic number Z = 3) and Hydrogen (atomic number Z = 1) both have one electron, you can use the Rydberg equation for the atomic energy levels:

E(n; Z) = -R*(Z^2)/(n^2),

where where R is the Rydberg constant for the atom/ion, Z is the atomic number of the element (the number of protons in the nucleus), and n is the quantum number for that energy level (n = 1, 2, 3, ...). Every element has its own Rydberg constant, but they're all pretty close to 13.6 eV = 2.18 x 10^-18 Joule (1 eV = 1.6 x 10^-19 J). (This energy is relative to E = 0 when the electron becomes completely dissociated from this ion.)

The first excited state of hydrogen is n = 2 (n = 1 is the lowerst-energy, or "ground" state), which from the Rydberg equation has an energy equal to E(n = 2; Z = 1) = -R/4 = -3.4 eV.

The states of Lithium 2+ have energies equal to E(n, Z = 3) = -9R/(n^2). To solve for n (which tells us what excited state the ion is in), we need to equate the hydrogen first-excited state energy with this energy:

-R/4 = -9R/(n^2).

Solving for n gives:

n = sqrt(9 x 4) = sqrt(36) = 6.

So, the n = 6 (fifth excited) state of Li 2+ has the same energy as the n = 2 (first excited) state of hydrogen, according to the Rydberg equation for hydrogen-like (one-electron) atoms.