First line in balmer series means transition from $\mathrm{n}=3$ to $\mathrm{n}=2$. Now, $9^{\mathrm{t}\mathrm{h}}$ orbit of ${\mathrm{L}\mathrm{i}}^{2+}$ has same energy as third orbit of $\mathrm{\prime }\mathrm{H}\mathrm{\prime }$ atom & $6^{\mathrm{t}\mathrm{h}}$ orbit of ${\mathrm{L}\mathrm{i}}^{2+}$ has same energy as second orbit of $\mathrm{\prime }\mathrm{H}\mathrm{\prime }$ atom. So $\mathrm{n}=9$ to $\mathrm{n}=6$ transition in ${\mathrm{L}\mathrm{i}}^{2+}$would have same wavelength as first line in balmer series of

$\mathrm{H}$-spectrum.