Electromagnetic radiation of wavelength 663 nm is just sufficient to ionise the atom of metal A . The ionization energy of metal $A$ in $\mathrm{kJ} \mathrm{mol}^{-1}$ is $\_\_\_\_$ (Rounded-off to the nearest integer) $$ \begin{aligned} & {\left[\mathrm{h}=6.63 \times 10^{-34} \mathrm{Js}, \mathrm{c}=3.00 \times 10^8 \mathrm{~ms}^{-1}\right.} \\ & \left.\mathrm{N}_{\mathrm{A}}=6.02 \times 10^{23} \mathrm{~mol}^{-1}\right] \end{aligned} $$