The given wave equation is $E = 40 \cos(kz - 6 \times 10^8 t)$. Comparing this with the standard equation of a plane electromagnetic wave propagating in the $z$-direction, $E = E_0 \cos(kz - \omega t)$, we identify the angular frequency $\omega = 6 \times 10^8 \text{ rad s}^{-1}$ . The speed of light in vacuum ($c$) relates the angular frequency ($\omega$) and the magnitude of the wave vector ($k$) by the relation: $c = \frac{\omega}{k}$ where $c = 3 \times 10^8 \text{ m s}^{-1}$. Rearranging for $k$: $k = \frac{\omega}{c} = \frac{6 \times 10^8}{3 \times 10^8} = 2 \text{ m}^{-1}$