Answer
$6.56\times10^{-7}\,m$ and $1.22\times10^{-7}\,m$
Work Step by Step
The atom makes transition from $n=3$ to $n=2$ at first.
Corresponding energy change
$\Delta E=-2.18\times10^{-18}\,J(\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}})=-2.18\times10^{-18}\,J(\frac{1}{2^{2}}-\frac{1}{3^{2}})$
$=-3.03\times10^{-19}\,J$
Negative sign indicates that the energy is emitted.
$\Delta E=\frac{hc}{\lambda}$
$\implies$ Wavelength $ \lambda=\frac{hc}{\Delta E}$ where $h$ is the Planck's constant and $c$ is the speed of light.
$\lambda =\frac{(6.626\times10^{-34}\,Js)(3.00\times10^{8}\,m/s)}{3.03\times10^{-19}\,J}=6.56\times10^{-7}\,m$
Then, atom makes transition from $n=2$ to $n=1$.
Corresponding energy change
$\Delta E=-2.18\times10^{-18}\,J(\frac{1}{1^{2}}-\frac{1}{2^{2}})=-1.635\times10^{-18}\,J$
$\lambda=\frac{hc}{\Delta E}=\frac{(6.626\times10^{-34}\,Js)(3.00\times10^{8}\,m/s)}{1.635\times10^{-18}\,J}=1.22\times10^{-7}\,m$
