Shortest wavelength means maximum energy in transition.

Therefore for Balmer series transition would be $n=\infty \text{to} n=2$ and for Brackett series transition would be from $n=\infty \text{to} n=4$.

Now from Rydberg's formula,

$$\begin{gathered} \frac{{\displaystyle \frac{1}{{\lambda }_{Bl}}}}{{\displaystyle \frac{1}{{\lambda }_{Br}}}}=\frac{R_H\left({\displaystyle \frac{1}{2^2}}-{\displaystyle \frac{1}{{\infty }^2}}\right)}{R_H\left(\frac{1}{4^2}-\frac{1}{{\infty }^2}\right)} \\ \Rightarrow {\lambda }_{Br}=4{\lambda }_{Bl}=4\lambda \end{gathered}$$