Answer
$9.65\times 10^{-20}$ $A$
Work Step by Step
If $P$ be the power of the laser, and $E$ be the energy of a single incident photon, the rate of incident photon can be expressed as:
$R_{in}=\frac{P}{E}$ ................$(1)$
Now, energy of a single photon is expressed as, $E=\frac{ch}{\lambda}$,
in which $c$ is the speed of light in vacuum, $\lambda$ is the wavelength of the light and $h$ is the Planck's constant
In our case, $\lambda=600$ $nm$ $=600\times 10^{-9}$ $m$, $P=2.00$ $mW$ $=2.00\times 10^{-3}$ $W$
Therefore, $E=\frac{3\times 10^{8}\times 6.63\times 10^{-34}}{600\times 10^{-9}}$ $J$
or, $E=3.315\times 10^{-19}$ $J$
$\therefore$ Using eq. $1$, the rate of incident photon is given by
$R_{in}=\frac{2.00\times 10^{-3}}{3.315\times 10^{-19}}$ $photons/s$
or, $R_{in}\approx 6.03\times10^{15}$ $photons/s$
The fractional efficiency of the cesium surface is $1.0\times 10^{-16}$ ; that is, on average one electron is ejected for every $10^{16}$ photons that reach the surface.
Therefore, the rate of electron emission is
$R_{em}=1.0\times 10^{-16}\times R_{in}$ $photons/s$
or, $R_{em}=1.0\times 10^{-16}\times 6.03\times10^{15}$ $photons/s$
or, $R_{em}=6.03\times 10^{-1}$ $photons/s$
Therefore, the current of electrons ejected from the surface is given by
$I=\text{Charge of an electron}\times R_{em}$
or, $I=1.6\times 10^{-19}\times 6.03\times 10^{-1}$ $A$
or, $I\approx 9.65\times 10^{-20}$ $A$
$\therefore$ The current of electrons ejected from the surface is $9.65\times 10^{-20}$ $A$
