Ratio of shortest wavelength of Balmer series in a hydrogen atom to that of Lymann series in a $H{e}^{+}$ ion is 

for any series shortest wavelength can be given by  $\frac{1}{\lambda }=R{Z}^2\left(\frac{1}{n^2}-\frac{1}{\infty }\right) where R is the Rydberg\text{'}s cons\tan t and Z is the atomic no.$

For Balmer series in hydrogen atom $\frac{1}{{\lambda }_1}=R\left(\frac{1}{2^2}-\frac{1}{\propto }\right)=\frac{R}{4}$

For Lymann series of $H{e}^{+}$ is $\frac{1}{{\lambda }_2}=4R\left(\frac{1}{1^2}-\frac{1}{\infty }\right)$

$\therefore \frac{{\lambda }_1}{{\lambda }_2}=\frac{4/R}{1/4R}=16$