A helium-neon laser sends out a beam of power 3.0 mW…

Step 1. The intensity. $r = 1.0$ mm, so $I = \dfrac{P}{\pi r^2} = \dfrac{0.0030\ \text{W}}{\pi \times (0.0010\ \text{m})^2} = 0.955\ \text{kW/m}^2$.

Step 2. The electric field. $E_0 = \sqrt{\dfrac{2I}{\varepsilon_0 c}} = \sqrt{\dfrac{2 \times 954.9}{2.655 \times 10^{-3}}}\ \text{V/m} = 848\ \text{V/m}$.

Step 3. The magnetic field. $B_0 = \dfrac{E_0}{c} = \dfrac{848\ \text{V/m}}{3.00 \times 10^{8}\ \text{m/s}} = 2.83\ \mu\text{T}$.

Step 4. The energy in a metre. $U = \dfrac{PL}{c} = \dfrac{0.0030\ \text{W} \times 1.0\ \text{m}}{3.00 \times 10^{8}\ \text{m/s}} = 10.0\ \text{pJ}$.

Step 5. The photons. Here $hc = 1.989 \times 10^{-25}\ \text{J m}$. One photon has $\dfrac{hc}{\lambda} = \dfrac{1.989 \times 10^{-25}\ \text{J m}}{632.8 \times 10^{-9}\ \text{m}} = 3.14 \times 10^{-19}\ \text{J}$. So $N = \dfrac{10.0 \times 10^{-12}\ \text{J}}{3.14 \times 10^{-19}\ \text{J}} = 3.18 \times 10^{7}$.