The wavelength of the first spectral line of the Lyman series of hydrogen spectrum is

$\frac{1}{\lambda }=R\left(\frac{1}{{\left(n_1\right)}^2}-\frac{1}{{\left(n_2\right)}^2}\right)\cdot Z^2$

where,

$R=$ Rydbergs constant (Also written is $R_H$)

$Z=$ atomic number

Since the question is asking for $1^{\mathrm{st}}$ line of Lyman series therefore

$n_1=1$

$n_2=2$

Since the electron is de-exited from $1\left(\mathrm{st}\right)$ exited sate $\left(\mathrm{i}.\mathrm{e}.n=2\right)$ to ground sate $\left(\mathrm{i}.\mathrm{e}.n=1\right)$ for first line of Lyman series.

$\frac{1}{\lambda }=R\left(\frac{1}{{\left(1\right)}^2}-\frac{1}{{\left(2\right)}^2}\right)\cdot 1^2$

Since the atomic number of Hydrogen is $1$

By doing the math, we get the wavelength as

$\lambda =\frac{4}{3}\cdot 912 \mathrm{Å}$

Since $\frac{1}{R}=912 \mathrm{Å}$

therefore

$\lambda =1216 \mathrm{Å}$