Answer
number of photons per second per square meter that reach earth will be $2.495688\times10^{21}$.
Work Step by Step
Intensity of light is given as $I= 680W/m^2=680J/(s.m^2)$
it means there is incident energy of $E=680 J$ per second per square meter .........statement(1)
Given that all photon are of wavelength $\lambda=730nm=730\times10^{-9}m=7.3\times10^{-7}m$
From Planks hypothesis
$E=\frac{hc}{\lambda}$
putting $h=6.63\times10^{-34}J.s$, $\lambda=7.3\times10^{-7}m$,, $c=3\times10^{8}m/s$
so Energy of one photon
$E=\frac{hc}{\lambda}=\frac{6.63\times10^{-34}J.s\times3\times10^{8}m/s}{7.3\times10^{-7}m}=2.7247\times10^{-19}J$
Now since $2.7247\times10^{-19}J$ is equal to $1$ photon
$1J$ will be equal to $\frac{1}{2.7247\times10^{-19}}$photons
so $680J$ is equal to $\frac{680}{2.7247\times10^{-19}}$photons=$249.5688\times10^{19}$ photons
or $680J$ is equal to $2.495688\times10^{21}$ photons
Now statement (1) can be written as number of photons per second per square meter that reach earth will be $2.495688\times10^{21}$.
