The electric field in a plane electromagnetic wave is given by, $$E = 50\sin(500x - 10 \times 10^{10}t)$$ V m$$^{-1}$$. The velocity of an electromagnetic wave in this medium is: (Given $$c$$ = the speed of light in vacuum).

We observe that the electric field of the given plane electromagnetic wave is written in the form

$$E = 50 \sin(500x - 10 \times 10^{10} t) \; \text{V m}^{-1}$$

The standard mathematical expression for a monochromatic plane wave travelling along the positive $$x$$-direction is

$$E = E_0 \sin(kx - \omega t),$$

where $$k$$ is the angular wave number and $$\omega$$ is the angular frequency. By simply matching every symbol, we equate

$$k = 500 \; \text{rad m}^{-1}, \qquad \omega = 10 \times 10^{10} \; \text{rad s}^{-1}.$$

First, make the numerical value of $$\omega$$ explicit:

$$\omega = 10 \times 10^{10} = 1 \times 10^{11} \; \text{rad s}^{-1}.$$

The speed of a wave is connected to these parameters through the relation

$$v = \frac{\omega}{k}.$$

Substituting the recognised values, we write

$$v = \frac{1 \times 10^{11}}{500} \; \text{m s}^{-1}.$$

Now we carry out the division step by step. First, divide the numerator and denominator:

$$\frac{1 \times 10^{11}}{5 \times 10^{2}} = 0.2 \times 10^{9}$$

(because $$500 = 5 \times 10^{2}$$ and we moved one power of ten from the denominator to the numerator). Converting $$0.2 \times 10^{9}$$ into a more familiar decimal form, we obtain

$$0.2 \times 10^{9} = 2 \times 10^{8} \; \text{m s}^{-1}.$$

Next, we compare this value with the speed of light in vacuum, $$c = 3 \times 10^{8} \; \text{m s}^{-1}$$. Forming the ratio,

$$\frac{v}{c} = \frac{2 \times 10^{8}}{3 \times 10^{8}} = \frac{2}{3}.$$

So the speed of this electromagnetic wave in the medium is

$$v = \frac{2}{3}c.$$

Hence, the correct answer is Option D.