1. For a hydrogen-like species, the radius of the $n^{\text{th}}$ orbit is

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

where $a_0$ is the Bohr radius and $Z$ is the atomic number.

2. The difference between the radii of 4th and 3rd orbits is

$\Delta R = r_4 - r_3 = a_0\frac{4^2-3^2}{Z} = a_0\frac{16-9}{Z} = a_0\frac{7}{Z}$

So, for any hydrogen-like ion,

$\Delta R \propto \frac{1}{Z}$

3. For $Li^{2+}$, we have $Z=3$.

Hence,

$\Delta R_1 = a_0\frac{7}{3}$

4. For $He^+$, we have $Z=2$.

Hence,

$\Delta R_2 = a_0\frac{7}{2}$

5. Therefore,

$\Delta R_1 : \Delta R_2 = \frac{7a_0}{3} : \frac{7a_0}{2} = \frac{1}{3} : \frac{1}{2} = 2:3$

6. So the correct option is:

$\boxed{\text{C: } 2:3}$

7. Comparison with stored correct answer:

Stored correct answer = C, which matches our derived answer.