In a particular system of units, a physical quantity can be...

Given here: $\left[B\right] = {\left[e\right]}^{\alpha}{\left[{m}_{e}\right]}^{\beta}{\left[h\right]}^{\gamma}{\left[k\right]}^{\delta}$

The dimensions of \left[e\right] = \left[Current \times time\right] = \left[IT\right] and \left[{m}_{e}\right] = \left[M\right]$, Planck's constant \left[h\right] = \dfrac{Energy}{Frequency} = \dfrac{\left[M{L}^{2}{T}^{- 2}\right]}{\left[{T}^{- 1}\right]} = \left[M{L}^{2}{T}^{- 1}\right]$ and \left[k\right] = \dfrac{1}{\left[{ε}_{0}\right]} = {\left[{M}^{- 1}{L}^{- 3}{T}^{4}{I}^{2}\right]}^{- 1} = \left[M{L}^{3}{T}^{- 4}{I}^{- 2}\right]</p><p>Using Principle of $homogeneity,</p><p>\left[{M}^{1}{T}^{- 2}{I}^{- 1}\right] = {\left[IT\right]}^{\alpha}{\left[M\right]}^{\beta}{\left[M{L}^{2}{T}^{- 1}\right]}^{\gamma}{\left[M{L}^{3}{T}^{- 4}{I}^{- 2}\right]}^{\delta}$

Comparing both sides, we get

$\beta + \gamma + \delta = 1 . . . \left(i\right)</p><p>$2\gamma + 3\delta = 0 . . . \left(ii\right)$</p><p>$\alpha - \gamma - 4\delta = - 2 . . . \left(iii\right)

$\alpha - 2\delta = - 1 . . . \left(iv\right)$

On solving above four equations, we get

$\alpha = 3, \beta = - 4, \gamma = 3$ and $\delta = 2$

So, $\alpha + \beta + \gamma + \delta = 4$