Given that the area of the metal plate $$A = 1 \times 10^{-4} m^2$$, and the intensity of the radiation

$$I = 16 mW/m^2 = 16 \times 10^{-3} W/m^2$$.

The power $$P$$ falling on the metal plate is given by the product of the intensity of the radiation and the area:

$$P = IA = (16 \times 10^{-3} W/m^2)(1 \times 10^{-4} m^2) = 1.6 \times 10^{-6} W$$.

The energy of each incident photon is given as 10 eV. Converting this to joules (since 1 eV = $$1.6 \times 10^{-19} J$$):

$$E_{\text{photon}} = 10 eV = 10 \times 1.6 \times 10^{-19} J = 1.6 \times 10^{-18} J$$.

The number of incident photons per second $$n_{\text{photon}}$$ is then given by the total power divided by the energy per photon:

$$n_{\text{photon}} = P / E_{\text{photon}} = 1.6 \times 10^{-6} W / 1.6 \times 10^{-18} J = 10^{12} \text{ photons/s}$$.

Given that only 10% of these photons actually produce photoelectrons, the number of photoelectrons produced per second $$n_{\text{electron}}$$ is:

$$n_{\text{electron}} = 0.1 \times n_{\text{photon}} = 0.1 \times 10^{12} = 10^{11} \text{ electrons/s}$$.

The work function of the metal is given as 5 eV, which is the energy required to remove an electron from the metal. Therefore, the maximum energy of the photoelectrons $$E_{\text{max}}$$ is given by the energy of the incident photons minus the work function:

$$E_{\text{max}} = E_{\text{photon}} - \text{Work function} = 10 eV - 5 eV = 5 eV$$.