Thank you for the opportunity to help you with your question!

(a) To get the result for doubly ionized lithium, replace e^{2} by Ze^{2} in the expression for the energy levels.
E_{n} = (m_{e} k_{e}^{2} Z^{2} e^{4} / 2 hbar^{2}) (1/n^{2}) = -Z^{2} (13.6eV / n^{2}) = -122.4eV / n^{2}
The ground state energy is
E_{1} = -122.4eV

(b) Make the same change to the expression for the radius:
r_{n} = n^{2} hbar^{2} / m_{e} k_{e} Z e^{2} = n^{2} a_{0} / Z
The ground-state radius is
r_{1} = a_{0} / Z = 0.0529nm / 3 = 0.0176nm Please let me know if you need any clarification. I'm always happy to answer your questions.