What is the wave number of spectral line corresponding to second line of Balmer series of He+ ion?

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$\frac{1}{{\lambda }_1}=R(1-\frac{1}{2^2})=R(1-\frac{1}{4})=\frac{3R}{4}$
$\Rightarrow {\lambda }_1=\frac{4}{3R}$
$\therefore \frac{{\lambda }_1}{{\lambda }_2}=\frac{4}{3R}\times \frac{3R}{16}=\frac{1}{4}$
$\Rightarrow {\lambda }_1=\frac{{\lambda }_2}{4}=\frac{486.4}{4}=121.6nm$