Why is there a longer time lag if light behaves as a wave? In this problem:

According to a model based on the electromagnetic theory of light, the electron absorbs all the energy that is incident on the surface within a distance of $5.0\times 10^{-11}\ \mathrm{m}$ from the electron. The intensity of light incident on the surface is $1.6\ \mathrm{W\ m^{-2}}$. The energy required to remove an electron from the surface is $1.8\ \mathrm{eV}$.
(a) Calculate, on the basis of this model, that the time taken for the electron to gain sufficient energy to leave the surface is $23\ \mathrm{s}$. (The area of a circle of radius $R$ is $\pi R^2$.)
(b) Experimental observation indicates that electrons are emitted from the surface in less than $10^{-9}\ \mathrm{s}$. Explain how this observation is consistent with the particle theory of light.

The answer to the first part is trivial, however, my main problem lies in the second part. Although I've studied this concept several times I tend to forget it. And I believe that this is because, I've not really understood. As far as I can recall, if we assume that light behaves as a wave, then there will be a longer time lag. But why? 
 A: If you consider light to behave as particle then the energy is carried by the particles and when it collides with an electron the energy is transferred instantaneously.
But in case of wave the energy is distributed over the wave front and the time to transfer the same amount of energy will be longer.
A: As you have problems with the quantised picture of light, let us start from the quantised light picture and work toward the wave picture.


*

*Let us assume that we have photons (=quantised light particles). Furthermore, let us assume that each photon has the energy of 2eV, which is larger than the needed 1.8eV from your question. 

*What happens if an electron absorbs a photon? Well, the photon energy is transferred to the electron. As the electron gains enough energy to escape from the surface, it does so. Thus, "as soon as" the first photon hits the surface an electron is ejected from the surface. 

*Now let us try to work towards the wave picture, which considers the averaged energy of the photons. As each photons carries the energy of $2eV \approx 3.2 \cdot 10^{-19}J$ there are $1.6W / 2eV \approx 5 \cdot 10^{18}$ photons which hit the surface of $A=1m^2$ every second. However, in the wave picture we are allowed to transfer the absorption cross section from the electron to the photon. Hence, we ask what area do these photons cover, if each photon covers the area $\sigma = \pi R^2 \approx 7.8 \cdot 10^{-21}m^2$? Well, as we have $N=5 \cdot 10^{18}$ photons the total cross section is $N \cdot \sigma \approx 0.039m^2$. Therefore, on average the photons cover only 3.9% of the total area (which is $1m^2$). Thus they need 1/3.9% $\approx 25.5s$ to cover the total area. [we get a different number, because I used 2eV instead of 1.8eV].

*Next you should answer the question, what happens if we increase the photon energy? E.g. let's assume each photon carries the energy 10eV instead of 2eV. 

