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 $Z=2$