If electronic charge e , electron mass m , speed of light in vacuum...

Magnetic potential energy inside a solenoid is

$U = \dfrac{1}{2}L{I}^{2}$ Where $L =$self-inductor$⇒ U = \dfrac{1}{2}\left({\mu}_{0}{n}^{2}Al\right){I}^{2}$

$∵ n =$ number of turns per unit length

$= \dfrac{N}{l}$

$U = \dfrac{1}{2}{\mu}_{0}{\left(\dfrac{N}{l}\right)}^{2}Al{I}^{2}$

$U = \dfrac{1}{2}{\mu}_{0}\dfrac{{N}^{2}}{{l}^{2}}Al{I}^{2}$

${\mu}_{0} = \dfrac{2Ul}{{N}^{2}A{I}^{2}}$

Dimensionally, \left[U\right] = \left[\dfrac{hc}{\lambda}\right]$<br /><br />\left[I\right] = \left[\dfrac{e}{T}\right]$

\left[A\right] = {\left[l\right]}^{2}$<br /><br />\left[l\right] = \left[{l}^{′}\right]$

\left[{\mu}_{0}\right] = \left[\dfrac{hc}{\lambda}\right]\dfrac{[ l ]}{\left[{l}^{2}\right]}\dfrac{\left[{t}^{2}\right]}{{\left[e\right]}^{2}}$<br /><br />$= \left[\dfrac{h}{\lambda}\dfrac{t}{{e}^{2}}\right] = \left[\dfrac{h}{c{e}^{2}}\right]$<br /><br />${\mu}_{0} = \dfrac{h}{c{e}^{2}}