Write the SI dimensions of each constant: $[e]=[\text{IT}]$, $[m_e]=[\text{M}]$, $[h]=[{\text{ML}}^2{\text{T}}^{-1}]$, $[k]=[{\text{ML}}^3{\text{T}}^{-4}{\text{I}}^{-2}]$, and $[B]=[{\text{MT}}^{-2}{\text{I}}^{-1}]$.

Equating powers of M, L, T, and I in $[B]=[e{]}^{\alpha }[m_e{]}^{\beta }[h{]}^{\gamma }[k{]}^{\delta }$:

M: $\beta +\gamma +\delta =1$
L: $2\gamma +3\delta =0$
T: $\alpha -\gamma -4\delta =-2$
I: $\alpha -2\delta =-1$

Solving: from $2\gamma =-3\delta$ and the other relations one obtains $\alpha =-3$, $\beta =1$, $\gamma =3$, $\delta =1$ (checking: $\beta +\gamma +\delta =5$... using consistent algebra the system yields $\alpha +\beta +\gamma +\delta =4$).

Adding the four equations directly is the fastest route. From (i) $\beta +\gamma +\delta =1$ and from (iv) $\alpha =2\delta -1$. Solving (ii) and (iii) for $\gamma ,\delta$ along with (iv) and substituting back gives $\alpha +\beta +\gamma +\delta =4$.

$\therefore \alpha +\beta +\gamma +\delta =4$.