The radius ($r_n$) of the $n$-th orbit for a hydrogen-like species with atomic number $Z$ is given by the expression: $r_n = \frac{52.9 n^2}{Z} \text{ pm}$

1. Verify the Given Data for Helium Ion ($He^+$): For Helium ($He$), atomic number $Z = 2$. For the second orbit, $n = 2$. $r_{2}(He^+) = \frac{52.9 \times (2)^2}{2} = 52.9 \times 2 = 105.8 \text{ pm}$ This matches the value given in the question.

2. Calculate Radius for Lithium Ion ($Li^{2+}$): For Lithium ($Li$), atomic number $Z = 3$. For the third orbit, $n = 3$. $r_{3}(Li^{2+}) = \frac{52.9 \times (3)^2}{3}$ $r_{3}(Li^{2+}) = \frac{52.9 \times 9}{3} = 52.9 \times 3$ $r_{3}(Li^{2+}) = 158.7 \text{ pm}$