A photon of wavelength 6630 $\stackrel{\mathrm{o}}{\mathrm{A}}$ is incident on a totally reflecting surface. Momentum delivered by the photon is equal to 

Momentum of the incident photon is $\mathrm{P}=\frac{\mathrm{h}}{\mathrm{\lambda }}$, 
As it is a totally reflecting surface the change in momentum of the photon is 2P
So, $\mathrm{\Delta P}=2\mathrm{P}=2\frac{\mathrm{h}}{\mathrm{\lambda }}=2\frac{6.63\times {10}^{-34}}{6630\times {10}^{-10}}=2\times {10}^{-27}$kg m / s