From the following combinations of physical constants (expressed through their usual symbols) the only combination, that would have the same value in different systems of units, is:

To determine which combination of physical constants has the same value in different systems of units, we need to find the dimensionless quantity. A dimensionless quantity has dimensions of mass, length, time, and current all raised to the power zero, i.e., $$[M^0 L^0 T^0 I^0]$$. We will perform dimensional analysis on each option using the following dimensions of the fundamental constants:

Speed of light, $$c$$: $$[L T^{-1}]$$

Planck's constant, $$h$$: $$[M L^2 T^{-1}]$$ (since energy $$E = h\nu$$ and $$\nu$$ has $$[T^{-1}]$$)

Elementary charge, $$e$$: $$[I T]$$ (charge = current $$\times$$ time)

Gravitational constant, $$G$$: $$[M^{-1} L^3 T^{-2}]$$ (from Newton's law of gravitation)

Permittivity of free space, $$\varepsilon_0$$: $$[M^{-1} L^{-3} T^4 I^2]$$ (from Coulomb's law: $$F = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2}$$)

Permeability of free space, $$\mu_0$$: $$[M L T^{-2} I^{-2}]$$ (from the force between parallel currents)

Note that $$\mu_0 \varepsilon_0 = \frac{1}{c^2}$$, so the dimensions of $$\mu_0 \varepsilon_0$$ are $$[T^2 L^{-2}]$$.

Also, constants like $$2\pi$$ are dimensionless and do not affect the dimensional analysis.

Now, we evaluate each option:

Option A: $$\frac{ch}{2\pi\varepsilon_0^2}$$

Dimensions of numerator $$c h$$: $$[L T^{-1}] \times [M L^2 T^{-1}] = [M L^3 T^{-2}]$$

Dimensions of denominator $$\varepsilon_0^2$$: $$[M^{-1} L^{-3} T^4 I^2]^2 = [M^{-2} L^{-6} T^8 I^4]$$

Dimensions of fraction: $$\frac{[M L^3 T^{-2}]}{[M^{-2} L^{-6} T^8 I^4]} = M^{1 - (-2)} L^{3 - (-6)} T^{-2 - 8} I^{-4} = [M^3 L^9 T^{-10} I^{-4}]$$

This is not dimensionless.

Option B: $$\frac{e^2}{2\pi\varepsilon_0 G m_e^2}$$

Dimensions of numerator $$e^2$$: $$[I T]^2 = [I^2 T^2]$$

Dimensions of denominator $$\varepsilon_0 G m_e^2$$: $$\varepsilon_0: [M^{-1} L^{-3} T^4 I^2]$$, $$G: [M^{-1} L^3 T^{-2}]$$, $$m_e^2: [M]^2 = [M^2]$$, so $$\varepsilon_0 \times G \times m_e^2 = [M^{-1} L^{-3} T^4 I^2] \times [M^{-1} L^3 T^{-2}] \times [M^2] = M^{-1-1+2} L^{-3+3} T^{4-2} I^2 = [M^0 L^0 T^2 I^2]$$

Dimensions of fraction: $$\frac{[I^2 T^2]}{[T^2 I^2]} = [I^{2-2} T^{2-2}] = [M^0 L^0 T^0 I^0]$$

This is dimensionless.

Option C: $$\frac{\mu_0\varepsilon_0}{c^2} \cdot \frac{G}{h e^2}$$

First part: $$\frac{\mu_0\varepsilon_0}{c^2}$$, dimensions of $$\mu_0\varepsilon_0$$: $$[T^2 L^{-2}]$$, dimensions of $$c^2$$: $$[L^2 T^{-2}]$$, so fraction: $$\frac{[T^2 L^{-2}]}{[L^2 T^{-2}]} = T^{2 - (-2)} L^{-2 - 2} = [T^4 L^{-4}]$$

Second part: $$\frac{G}{h e^2}$$, dimensions of $$G$$: $$[M^{-1} L^3 T^{-2}]$$, dimensions of $$h e^2$$: $$h: [M L^2 T^{-1}]$$, $$e^2: [I^2 T^2]$$, so $$h e^2 = [M L^2 T^{-1}] \times [I^2 T^2] = [M L^2 T I^2]$$, fraction: $$\frac{[M^{-1} L^3 T^{-2}]}{[M L^2 T I^2]} = M^{-1-1} L^{3-2} T^{-2-1} I^{-2} = [M^{-2} L T^{-3} I^{-2}]$$

Product: $$[T^4 L^{-4}] \times [M^{-2} L T^{-3} I^{-2}] = M^{-2} L^{-4+1} T^{4-3} I^{-2} = [M^{-2} L^{-3} T I^{-2}]$$

This is not dimensionless.

Option D: $$\frac{2\pi\sqrt{\mu_0\varepsilon_0}}{c e^2} \cdot \frac{h}{G}$$

Note that $$\sqrt{\mu_0\varepsilon_0} = \sqrt{\frac{1}{c^2}} = \frac{1}{c}$$, so the expression simplifies to $$\frac{2\pi \cdot \frac{1}{c}}{c e^2} \cdot \frac{h}{G} = \frac{2\pi}{c^2 e^2} \cdot \frac{h}{G}$$

Dimensions of denominator $$c^2 e^2$$: $$c^2: [L^2 T^{-2}]$$, $$e^2: [I^2 T^2]$$, so $$c^2 e^2 = [L^2 T^{-2}] \times [I^2 T^2] = [L^2 I^2]$$

Dimensions of $$\frac{h}{G}$$: $$h: [M L^2 T^{-1}]$$, $$G: [M^{-1} L^3 T^{-2}]$$, so fraction: $$\frac{[M L^2 T^{-1}]}{[M^{-1} L^3 T^{-2}]} = M^{1-(-1)} L^{2-3} T^{-1-(-2)} = [M^2 L^{-1} T]$$

Product: $$\frac{1}{[L^2 I^2]} \times [M^2 L^{-1} T] = [M^2 L^{-3} T I^{-2}]$$

This is not dimensionless.

Only Option B is dimensionless and thus has the same value in different systems of units. Hence, the correct answer is Option B.