In the given electromagnetic wave $$E_y = 600 \sin(\omega t - kx) \text{ Vm}^{-1}$$, intensity of the associated light beam is (in $$\text{W/m}^2$$) : (Given $$\epsilon_0 = 9 \times 10^{-12} \text{ C}^2 \text{ N}^{-1} \text{ m}^{-2}$$)

We are asked to find the intensity of an electromagnetic wave described by $$E_y = 600 \sin(\omega t - kx) \, \text{Vm}^{-1}$$.

The average intensity of an electromagnetic wave is related to the amplitude of the electric field by the formula

$$I = \frac{1}{2} \epsilon_0 c E_0^2$$

Here, $$\epsilon_0$$ is the permittivity of free space, $$c$$ is the speed of light, and $$E_0$$ is the peak electric field amplitude.

From the given wave equation, the amplitude is $$E_0 = 600 \, \text{V/m}$$. The standard values are $$\epsilon_0 = 9 \times 10^{-12} \, \text{C}^2\text{N}^{-1}\text{m}^{-2}$$ and $$c = 3 \times 10^8 \, \text{m/s}$$.

Substituting these into the intensity formula gives

$$I = \frac{1}{2} \times 9 \times 10^{-12} \times 3 \times 10^8 \times (600)^2$$

First, compute $$\epsilon_0 c$$:

$$\epsilon_0 \times c = 9 \times 10^{-12} \times 3 \times 10^8 = 27 \times 10^{-4} = 2.7 \times 10^{-3}$$

Next, evaluate $$E_0^2$$:

$$E_0^2 = 600^2 = 3.6 \times 10^5$$

Therefore, the intensity is

$$I = \frac{1}{2} \times 2.7 \times 10^{-3} \times 3.6 \times 10^5 = \frac{1}{2} \times 972 = 486 \, \text{W/m}^2$$

The correct answer is Option D: 486 W/m^2.