If wavelength of the first line of the Paschen series of hydrogen...

Paschen series are spectral lines coming to $n = 3$

${1}^{st}$ line of Paschen series makes a transition from, $4 \to 3$

So, ${2}^{nd}$ line of Paschen series makes a transition from, $5 \to 3$

We know that,

$\dfrac{1}{\lambda} = R\left[1{n}_{1}^{2} - 1{n}_{2}^{2}\right]{Z}^{2}$

For ${1}^{st}$ line, $\dfrac{1}{\lambda} = R\left[\dfrac{1}{( 3{)}^{2}} - \dfrac{1}{( 4{)}^{2}}\right]{Z}^{2}$

and for ${2}^{nd}$ line,

$\dfrac{1}{{\lambda}^{'}} = R\left[\dfrac{1}{( 3{)}^{2}} - \dfrac{1}{( 5{)}^{2}}\right]{Z}^{2}$

or $\dfrac{\left(\dfrac{1}{\lambda}\right)}{\left(\dfrac{1}{{\lambda}^{'}}\right)} = \dfrac{R\left[\dfrac{7}{9 \times 16}\right]{Z}^{2}}{R\left[\dfrac{16}{9 \times 25}\right]{Z}^{2}}$

Given, $\lambda = 720nm$

or $\dfrac{\left(\dfrac{1}{720}\right)}{\left(\dfrac{1}{\lambda '}\right)} = \dfrac{175}{256}$

or $\dfrac{{\lambda}^{'}}{720} = \dfrac{175}{256}$

or ${\lambda}^{'} \approx 492nm$