In the Bohr model for a hydrogen-like atom, the radius of the $ n^{\text{th}} $ orbit is given by

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

where:

$ n $ is the principal quantum number,

$ a_0 $ is the Bohr radius, and

$ Z $ is the atomic number (the effective nuclear charge).

For the $ 5^{\text{th}} $ orbit ($ n = 5 $):

For $ \mathrm{Li}^{2+} $ (where $ Z = 3 $):

$$ r_5(\mathrm{Li}^{2+}) = \frac{5^2 a_0}{3} = \frac{25a_0}{3} $$

For $ \mathrm{He}^{+} $ (where $ Z = 2 $):

$$ r_5(\mathrm{He}^{+}) = \frac{5^2 a_0}{2} = \frac{25a_0}{2} $$

Now, the ratio of the radius of the 5th orbit of $ \mathrm{Li}^{2+} $ to that of $ \mathrm{He}^{+} $ is:

$$ \frac{r_5(\mathrm{Li}^{2+})}{r_5(\mathrm{He}^{+})} = \frac{\frac{25a_0}{3}}{\frac{25a_0}{2}} = \frac{25a_0}{3} \times \frac{2}{25a_0} = \frac{2}{3} $$

Thus, the ratio is $ \frac{2}{3} $.

The correct answer is Option B.