The difference between the radii of 3rd and 4th orbits of $$\text{Li}^{2+}$$ is $$\Delta R_1$$. The difference between the radii of 3rd and 4th orbits of $$\text{He}^+$$ is $$\Delta R_2$$. Ratio $$\Delta R_1 : \Delta R_2$$ is:

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

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

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

We first apply this formula to the ion $$\text{Li}^{2+}$$. For lithium, $$Z=3$$. Therefore

$$r_n(\text{Li}^{2+})=\frac{n^{2}a_0}{3}.$$

Putting $$n=4$$, we get

$$r_4(\text{Li}^{2+})=\frac{4^{2}a_0}{3}=\frac{16a_0}{3}.$$

Putting $$n=3$$, we get

$$r_3(\text{Li}^{2+})=\frac{3^{2}a_0}{3}=\frac{9a_0}{3}=3a_0.$$

Now the required difference for lithium is

$$\Delta R_1=r_4-r_3=\frac{16a_0}{3}-3a_0=\frac{16a_0}{3}-\frac{9a_0}{3}=\frac{7a_0}{3}.$$

Next we turn to the ion $$\text{He}^+$$. For helium, $$Z=2$$. The same formula gives

$$r_n(\text{He}^+)=\frac{n^{2}a_0}{2}.$$

Putting $$n=4$$, we get

$$r_4(\text{He}^+)=\frac{4^{2}a_0}{2}=\frac{16a_0}{2}=8a_0.$$

Putting $$n=3$$, we get

$$r_3(\text{He}^+)=\frac{3^{2}a_0}{2}=\frac{9a_0}{2}=4.5a_0.$$

The required difference for helium is therefore

$$\Delta R_2=r_4-r_3=8a_0-4.5a_0=3.5a_0=\frac{7a_0}{2}.$$

To find the desired ratio, we write

$$\frac{\Delta R_1}{\Delta R_2}=\frac{\dfrac{7a_0}{3}}{\dfrac{7a_0}{2}}=\frac{7a_0}{3}\times\frac{2}{7a_0}=\frac{2}{3}.$$

Hence

$$\Delta R_1:\Delta R_2=2:3.$$

Hence, the correct answer is Option C.