Answer
3 photons.
Work Step by Step
Divide the minimum energy by the energy of a single 550-nm photon, to find the number of photons.
$$E=nhf=E_{min}$$
$$n=\frac{E_{min}}{hf}=\frac{E_{min}\lambda}{hc}$$
$$n=\frac{(10^{-18}J)(550\times10^{-9}m)}{( 6.63\times10^{-34}J \cdot s)(3.00\times10^8m/s)}$$
$$n=2.77\approx 3 photons$$
