If electronic charge $$e$$, electron mass $$m$$, speed of light in vacuum $$c$$ and Planck's constant $$h$$ are taken as fundamental quantities, the permeability of vacuum $$\mu_0$$ can be expressed in units of:

We have to express the permeability of free space $$\mu_0$$ in terms of the four selected fundamental quantities: electronic charge $$e$$, electron mass $$m$$, speed of light in vacuum $$c$$ and Planck’s constant $$h$$. (The mass $$m$$ will finally not enter the expression, but it is one of the allowed building blocks.) Our aim is purely dimensional, so any numerical factors such as $$2,\;4\pi,$$ or the dimension-less fine-structure constant $$\alpha$$ can be ignored because they do not change the units.

First, we recall the well-known relation from Maxwell’s theory that connects the speed of light with the vacuum permittivity $$\varepsilon_0$$ and the vacuum permeability $$\mu_0$$:

$$c=\frac{1}{\sqrt{\mu_0\varepsilon_0}}.$$

Rearranging this formula gives an explicit expression for $$\mu_0$$:

$$\mu_0=\frac{1}{\varepsilon_0\,c^{2}}.$$

So, once we succeed in writing $$\varepsilon_0$$ with the desired basic quantities, we can immediately obtain $$\mu_0$$.

Now, we involve the fine-structure constant $$\alpha$$, which by definition is dimensionless:

$$\alpha=\frac{e^{2}}{4\pi\varepsilon_0\hbar c},$$

where $$\hbar=\dfrac{h}{2\pi}$$ is the reduced Planck constant. Substituting $$\hbar=\dfrac{h}{2\pi}$$ in the definition of $$\alpha$$, we obtain

$$\alpha=\frac{e^{2}}{4\pi\varepsilon_0}\;\frac{2\pi}{h\,c} =\frac{e^{2}}{2\,\varepsilon_0\,h\,c}.$$

Isolating $$\varepsilon_0$$ from the above equation gives

$$\varepsilon_0=\frac{e^{2}}{2\alpha\,h\,c}.$$

Because $$\alpha$$ is dimensionless and the factor $$2$$ is a mere numerical constant, they have no bearing on dimensional analysis. Hence, for the purpose of units we may write simply

$$\varepsilon_0\sim\frac{e^{2}}{h\,c}.$$

Substituting this expression of $$\varepsilon_0$$ into the earlier formula $$\mu_0=\dfrac{1}{\varepsilon_0\,c^{2}}$$, we get

$$\mu_0=\frac{1}{\left(\dfrac{e^{2}}{h\,c}\right)\;c^{2}} =\frac{h\,c}{e^{2}\,c^{2}} =\frac{h}{c\,e^{2}}.$$

Thus, in terms of the chosen fundamental quantities, the permeability of vacuum carries the units of $$\displaystyle\frac{h}{c\,e^{2}}.$$ This matches Option D in the given list.

Hence, the correct answer is Option D.