To find the value of $$x$$, we need to first determine the expressions for the radii of the specified orbits for $$\mathrm{He}^{+}$$ and $$\mathrm{Be}^{3+}$$ according to Bohr's model. The radius of an orbit in a hydrogen-like atom (an atom with only one electron) is given by:

$$r_n = \frac{n^2 \cdot h^2 \cdot \epsilon_0}{\pi \cdot Z \cdot e^2 \cdot m_e}$$

Where:

$$r_n$$ is the radius of the nth orbit
$$n$$ is the principal quantum number (orbit number)
$$h$$ is the Planck's constant
$$\epsilon_0$$ is the vacuum permittivity
$$Z$$ is the atomic number (number of protons in the nucleus)
$$e$$ is the elementary charge
$$m_e$$ is the mass of the electron
$$\pi$$ is the mathematical constant pi

In this problem, we are looking at the 2nd orbit of $$\mathrm{He}^{+}$$ (which has an atomic number $$Z = 2$$) and the 4th orbit of $$\mathrm{Be}^{3+}$$ (which has an atomic number $$Z = 4$$). Let's calculate the radii for these orbits:

For the 2nd orbit of $$\mathrm{He}^{+}$$ ($$n_1 = 2$$ and $$Z_1 = 2$$):

$$r_{1} = \frac{n_1^2 \cdot h^2 \cdot \epsilon_0}{\pi \cdot Z_1 \cdot e^2 \cdot m_e}$$

For the 4th orbit of $$\mathrm{Be}^{3+}$$ ($$n_2 = 4$$ and $$Z_2 = 4$$):

$$r_{2} = \frac{n_2^2 \cdot h^2 \cdot \epsilon_0}{\pi \cdot Z_2 \cdot e^2 \cdot m_e}$$

We are asked to find the ratio $$\frac{r_{2}}{r_{1}}$$, which is equal to $$x: 1$$:

$$\frac{r_{2}}{r_{1}} = \frac{\frac{n_2^2 \cdot h^2 \cdot \epsilon_0}{\pi \cdot Z_2 \cdot e^2 \cdot m_e}}{\frac{n_1^2 \cdot h^2 \cdot \epsilon_0}{\pi \cdot Z_1 \cdot e^2 \cdot m_e}}$$

By simplifying the expression, we get:

$$\frac{r_{2}}{r_{1}} = \frac{n_2^2 \cdot Z_1}{n_1^2 \cdot Z_2} = \frac{4^2 \cdot 2}{2^2 \cdot 4}$$

Now we can calculate the value of $$x$$:

$$x = \frac{r_{2}}{r_{1}} = \frac{16 \cdot 2}{4 \cdot 4} = \frac{32}{16} = 2$$

Therefore, the value of $$x$$ in the ratio $$\frac{r_{2}}{r_{1}} = x: 1$$ is $$\boxed{2}$$.

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