To solve this problem, we will use the formula for the radius of the $n$-th Bohr orbit:$\\ r_n = \frac{n^2 h^2}{4 \pi^2 m e^2 Z} \\$where:$\\$• $n$ is the principal quantum number (orbit number),$\\$• $h$ is Planck's constant,$\\$• $m$ is the electron mass,$\\$• $e$ is the elementary charge,$\\$• $Z$ is the atomic number of the ion.$\\$The radius of the $n$-th orbit can also be expressed as:$\\ r_n = \frac{n^2 a_0}{Z} \\$where $a_0$ is the Bohr radius (approximately $52.9$ pm).$\\$For the He$^+$ ion in the second orbit:$\\ r_2 = \frac{2^2 a_0}{2} = 105.8$ pm$\\$We can solve for $a_0: \\ a_0 = \frac{105.8 \times 2}{4} \\ a_0 = 52.9$ pm$\\$Now, for the Li$^{2+}$ ion in the third orbit:$\\ r_3 = \frac{3^2 a_0}{3} \\ r_3 = \frac{9 \times 52.9}{3} \\ r_3 = 158.7$ pm$\\$Therefore, the radius of the third Bohr orbit of the Li$^{2+}$ ion is $158.7$ pm.$\\$This corresponds to Option 1.