The radius of a hydrogen-like ion according to Bohr's atomic model is given by the formula:

$ r = r_0 \frac{n^2}{Z} $

Where:

$ r_0 $ is the first Bohr radius ($ a_0 $)

$ n $ represents the principal quantum number

$ Z $ is the atomic number of the ion

For the ion $ \text{Li}^{++} $, the atomic number $ Z = 3 $ and we are considering the ground state, so $ n = 1 $.

Plugging these values into the formula, we get:

$ r = r_0 \frac{1^2}{3} = \frac{r_0}{3} $

This shows that the radius of $\text{Li}^{++}$ in its ground state is $\frac{1}{3} a_0$, meaning $ X = 3 $.