Answer
486 nm, visible region
Work Step by Step
$\Delta E_{atom}=-2.18\times10^{-18}\,J(\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}})$
$=-2.18\times10^{-18}\,J(\frac{1}{2^{2}}-\frac{1}{4^{2}})$
$=-4.0875\times10^{-19}\,J$
$E_{photon}=-\Delta E_{atom}=4.0875\times10^{-19}\,J$
$E=\frac{hc}{\lambda}\implies \lambda=\frac{hc}{E}$
$=\frac{(6.626\times10^{-34}\,J\cdot s)(3.00\times10^{8}\,m/s)}{4.0875\times10^{-19}\,J}$
$=4.86\times10^{-7}\,m=486\,nm$
This light is found in the visible region (In the visible region, wavelength ranges from 400 nm to 700 nm).