For ${\mathrm{He}}^{+}$ ion, the wave number associated with the Balmer transition, $\mathrm{n} = 4$ to $\mathrm{n} = 2$ is given by:

$\frac{1}{\mathrm{\lambda }} = {\mathrm{RZ}}^2 \left(\frac{1}{{{\mathrm{n}}_1}^2} - \frac{{\displaystyle 1}}{{\displaystyle {{\mathrm{n}}_2}^2}}\right)$

Where, ${\mathrm{n}}_1 = 2$

${\mathrm{n}}_2 = 4$

$\mathrm{Z}$ = atomic number of helium

$\frac{1}{\lambda }=\mathrm{R}(2{)}^2\left[\frac{1}{2^2}-\frac{1}{4^2}\right]$

$\Rightarrow \frac{1}{\mathrm{\lambda }}=4\mathrm{R}\left[\frac{4-1}{16}\right]$

$\Rightarrow \frac{1}{\mathrm{\lambda }}= \frac{3\mathrm{R}}{4}$

$\Rightarrow \mathrm{\lambda } = \frac{4}{3\mathrm{R}}$

According to the question, the desired transition for hydrogen will have the same wavelength as that of ${\mathrm{He}}^{+}$.

$\lambda \left(\mathrm{H}\right)=\lambda \left({\mathrm{He}}^{+}\right)$

$\mathrm{R}(1{)}^2\left[\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2}\right] = \frac{3\mathrm{R}}{4}$

By hit and trial method, 

$\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2}=\frac{1}{1^2}-\frac{1}{2^2}$

On comparing ${\mathrm{n}}_1=1 \& {\mathrm{n}}_2=2$

So, the correct answer is n = 2 to n = 1.