write $\mathfrak{B}\mathcal{O}_L$ as $\pi_L^{-a}\mathcal{O}_L$ for a certain integer $a {\geqslant}0$. Let now $x$ be a generator of $\mathfrak{B}  \mathcal{O}_L$ over $\mathfrak{B}$ (in particular, $x$ is a normal basis generator for $L/K$), and let $0\neq r\in\mathcal O_K$ be such that $rx\in \mathcal{O}_L$. Then we have: $$\begin{split}
  [\mathfrak{B}  \mathcal O_L : \mathcal O_L]  [\mathcal O_L : \mathcal O_K[G] rx]&=[\mathfrak{B}  \mathcal O_L : \mathcal O_K[G] rx]\\
  &=[\mathfrak{B}x : \mathcal O_K[G] x]  [\mathcal O_K[G] x : \mathcal O_K[G] rx],
\end{split}$$ and since $x$ is a normal basis generator this quantity is equal to $$\begin{split} 
[\mathfrak{B} : \mathcal O_K[G] ]\cdot [\mathcal O_K[G]  : \mathcal O_K[G] r]   & =[\mathfrak{B} : \mathcal O_K[G] ]\cdot |\mathcal{O}_K/r\mathcal{O}_K|^{|G|} \\
  & =[\mathfrak{B} : \mathcal O_K[G] ]\cdot p^{{f_K|G|v_K(r)}},
 \end{split}$$ where we have used that $\mathcal O_K[G]$ is a free $\mathcal{O}_K$-module of rank $|G|$. Hence we have $$m(L/K) {\leqslant}\left[  \mathfrak{B} : \mathcal{O}_K[G] \right] \cdot \frac{p^{{f_K|G|v_K(r)}}}{[ \mathfrak{B}   \mathcal{O}_L : \mathcal{O}_L]},$$ and we may replace $[ \mathfrak{B}   \mathcal{O}_L : \mathcal{O}_L]$ by $[ \pi_L^{-a} \mathcal{O}_L : \mathcal{O}_L] = p^{a f_L}$. Given that $\mathfrak{B}\mathcal{O}_L$ is the fractional ideal $(\pi_L^{-a})$, it is clear that $\pi_K^b x$ is in $\mathcal{O}_L$ provided that $v_L(\pi_K^b) {\geqslant}a$, that is, $
b {\geqslant}\frac{a}{e_{L/K}}.
$ Plugging in $r=\pi_K^b$ (with $b=\lceil\frac{a}{e_{L/K}}\rceil$) in the formula above, 