Answer
$3.47\times10^{52}\;\rm photons/year$
Work Step by Step
First of all, we need to find the area in which we need to count the photons that fall on it.
We start from the Sun and end at the Earth which is a spherical area that has a radius of the distance between the Sun and the Earth.
$$A=4\pi R^2_{\rm S\;to\;E}$$
Now we need to find the Sun's full power, which is given by
$$P=IA$$
where $I$ is the intensity, so
$$P=4\pi I R^2_{\rm S\;to\;E}$$
Recall that the energy produced by the sun to apply this power is given by the power times the time. So we need to find the energy produced by the sun in one year to give an intensity of 1350 W/m$^2$ on Earth's surface.
$$\Delta E=P\Delta t$$
Thus,
$$\Delta E=4\pi I \Delta t R^2_{\rm S\;to\;E}$$
And to find the number of photons,
$$N=\dfrac{\Delta E}{E_{p}}$$
where $E_p$ is the energy of one photon which is given by $E_p=hf=\dfrac{hc}{\lambda}$
Thus,
$$N=\dfrac{\Delta E\lambda}{hc } $$
$$N=\dfrac{4\pi I \Delta t \lambda R^2_{\rm S\;to\;E} }{hc } $$
Plugging the known;
$$N=\dfrac{4\pi \times 1350\times (365.25\times24\times 60^2) (550\times10^{-9}) (150\times10^9)^2 }{6.626\times10^{-34}\times3\times10^8} $$
$$N=\color{red}{\bf 3.47\times10^{52}}\;\rm photons/year$$
