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Given:
Wavelength of radiation, $\lambda = 6600\text{ \AA} = 6600 \times 10^{-10}\text{ m} = 6.6 \times 10^{-7}\text{ m}$
Incident power (energy per second), $P = 3 \times 10^{-8}\text{ W}$
Planck's constant, $h = 6.626 \times 10^{-34}\text{ J}\cdot\text{s}$ (or $6.63 \times 10^{-34}\text{ J}\cdot\text{s}$)
Speed of light, $c = 3 \times 10^8\text{ m/s}$

Energy of each photon:
$E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{6.6 \times 10^{-7}} = 3.012 \times 10^{-19}\text{ J}$

Number of photons incident per second:
$n = \frac{P}{E} = \frac{3 \times 10^{-8}\text{ W}}{3.012 \times 10^{-19}\text{ J}} \approx 9.96 \times 10^{10}\text{ s}^{-1} \approx 1.0 \times 10^{11}\text{ s}^{-1}$

Since each photon causes emission of one photoelectron (quantum efficiency is 100%), the number of photo-electrons emitted per second is:
$N = n \approx 9.96 \times 10^{10} \text{ (or } 1.0 \times 10^{11}\text{)}$