Use the hydrogen spectral series formula

For any spectral line, $\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right), \qquad n_2>n_1$

Identify the lowest wavelength of the Lyman series

In the Lyman series, transitions end at $n_1=1$.

The lowest wavelength means the series limit, i.e. $n_2\to\infty$.

So, $\frac{1}{\lambda_L}=R\left(1-0\right)=R$ Hence, $\lambda_L=\frac{1}{R}$

Given: $\lambda_L=917\,\text{\AA}$

Find the lowest wavelength of the Balmer series

In the Balmer series, transitions end at $n_1=2$.

Again, the lowest wavelength is the series limit, i.e. $n_2\to\infty$.

Therefore, $\frac{1}{\lambda_B}=R\left(\frac{1}{2^2}-0\right)=\frac{R}{4}$

So, $\lambda_B=\frac{4}{R}=4\lambda_L$

Substitute the given value

$\lambda_B=4\times 917=3668\,\text{\AA}$

Final Answer

$\boxed{3668\,\text{\AA}}$