If the series limit of the Lyman series for the hydrogen emission...

The wavelength of a spectral line in the hydrogen emission spectrum is given by the Rydberg formula:

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

For the series limit of the Lyman series, the transition is from $n_2 = \infty$ to $n_1 = 1$.

$\dfrac{1}{\lambda} = R \left( \dfrac{1}{1^2} - \dfrac{1}{\infty^2} \right) = R$

For the first line of the Balmer series, the transition is from $n_2 = 3$ to $n_1 = 2$. Let its wavelength be $\lambda'$.

$\dfrac{1}{\lambda'} = R \left( \dfrac{1}{2^2} - \dfrac{1}{3^2} \right)$

$\dfrac{1}{\lambda'} = R \left( \dfrac{1}{4} - \dfrac{1}{9} \right) = R \left( \dfrac{5}{36} \right)$

Substituting $R = \dfrac{1}{\lambda}$:

$\dfrac{1}{\lambda'} = \dfrac{1}{\lambda} \left( \dfrac{5}{36} \right)$

$\lambda' = \dfrac{36}{5}\lambda$

Answer: $\dfrac{36}{5}\lambda$