The radius of the second Bohr orbit, in terms of the Bohr radius, $$a_0$$, in Li$$^{2+}$$ is:

For any one-electron species, Bohr’s model gives the radius of the $$n^{\text{th}}$$ orbit as

$$r_n=\frac{n^2h^2\varepsilon_0}{\pi m_e e^2Z}.$$

In hydrogen, where $$Z=1$$ and $$n=1$$, this radius is called the Bohr radius $$a_0$$, so we write

$$a_0=\frac{h^2\varepsilon_0}{\pi m_e e^2}.$$

Dividing the two expressions term by term, the constant factors cancel, and we obtain the general relation

$$r_n=\frac{n^2a_0}{Z}.$$

For the doubly-ionised lithium ion, $$\text{Li}^{2+}$$, the nuclear charge is $$Z=3$$ because the nucleus has three protons. The question asks for the radius of the second Bohr orbit, so we put $$n=2$$. Substituting these values into the formula just derived gives

$$r_2=\frac{(2)^2a_0}{3}=\frac{4a_0}{3}.$$

This matches Option C.

Hence, the correct answer is Option C.