So I have a question like this :

Cesium has a work function of 1.8 eV. When light of wavelength 400 nm falls on the cathode of a photocell of area 1.2 cm^2, only 1 out of every 5 photons succeeded in ejecting an electron. The photoelectric current is 0.25 $\displaystyle \mu A$ Calculate the :

(i) Rate of emission of the photoelectrons from the cathode

(ii) Intensity of the incident light

I know the formula of intensity of incident light is $\displaystyle \frac{N}{t}$ $\displaystyle \frac{hc}{\lambda} \frac{1}{Area}$ But i don't know how to apply it here

Please help me..

Thankss