The shortest wavelength of Brackett series of a hydrogen like atom (atomic number$=Z$) is the same as the shortest wavelength of Balmer series of hydrogen atom. The value of $Z$ is:

Shortest wavelength of Brackett series corresponds to the transition of electron between $n_1=4$ and $n_2=\infty$ and the shortest wavelength of Balmer series corresponds to the transition of electron between $n_1=2$ and $n_2=\infty ,$ So,

                        $\left(Z^2\right)\left(\frac{13.6}{16}\right)=\left(\frac{13.6}{4}\right)$

$\therefore Z^2=4$  or $Z=2$