Assuming the sun is directly overhead, the surface of the earth...

The intensity of sunlight at the earth's surface is given by $I = 1.4 \times 10^3 \text{ W m}^{-2}$.

The energy of a single photon is given by $E = \dfrac{hc}{\lambda}$.

The number of photons crossing a unit area per unit time is $n = \dfrac{I}{E} = \dfrac{I \lambda}{hc}$.

Since these photons travel at the speed of light $c$, the number of photons present per unit volume (photon density) is $N = \dfrac{n}{c} = \dfrac{I \lambda}{hc^2}$.

Substituting the given values:

$N = \dfrac{1.4 \times 10^3 \times 500 \times 10^{-9}}{6.63 \times 10^{-34} \times (3 \times 10^8)^2}$

$N = \dfrac{7 \times 10^{-4}}{6.63 \times 10^{-34} \times 9 \times 10^{16}}$

$N = \dfrac{7 \times 10^{-4}}{59.67 \times 10^{-18}}$

$N \approx 1.17 \times 10^{13} \text{ m}^{-3}$

Rounding to appropriate significant figures, the number of photons per cubic metre is approximately $1.2 \times 10^{13}$.