A hyperbola has its centre at the origin, passes through the point $$(4, 2)$$ and has transverse axis of length 4 along the $$x$$-axis. Then the eccentricity of the hyperbola is:

We are told that the centre of the hyperbola is the origin, its transverse axis lies along the $$x$$-axis and the length of this transverse axis is 4.

For a hyperbola whose transverse axis is along the $$x$$-axis with centre at the origin, the standard equation is

$$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1.$$

Here $$2a$$ represents the length of the transverse axis. Since the given length is 4, we have

$$2a = 4.$$

Dividing both sides by 2, we get

$$a = 2.$$

So the equation of the required hyperbola can now be written as

$$\frac{x^{2}}{2^{2}}-\frac{y^{2}}{b^{2}}=1,$$

which simplifies to

$$\frac{x^{2}}{4}-\frac{y^{2}}{b^{2}}=1.$$

The hyperbola passes through the point $$(4,\,2)$$. That means when $$x = 4$$ and $$y = 2$$, the left side of the equation must equal 1. Substituting these values, we have

$$\frac{(4)^{2}}{4}-\frac{(2)^{2}}{b^{2}} = 1.$$

Calculating the squares gives

$$\frac{16}{4}-\frac{4}{b^{2}} = 1.$$

Simplifying the first fraction, we obtain

$$4-\frac{4}{b^{2}} = 1.$$

Now we isolate the term containing $$b^{2}$$ by subtracting 1 from both sides:

$$4-\frac{4}{b^{2}}-1 = 0 \quad\Longrightarrow\quad 3-\frac{4}{b^{2}} = 0.$$

Next, we add $$\dfrac{4}{b^{2}}$$ to both sides to remove the negative sign:

$$3 = \frac{4}{b^{2}}.$$

To solve for $$b^{2}$$, we take the reciprocal of both sides:

$$\frac{1}{3} = \frac{b^{2}}{4}.$$

Multiplying both sides by 4, we get

$$b^{2} = \frac{4}{3}.$$

We are now ready to compute the eccentricity $$e$$. For a hyperbola in the form

$$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}} = 1,$$

the formula for the eccentricity is

$$e = \sqrt{1+\frac{b^{2}}{a^{2}}}.$$

We already have $$a^{2}=4$$ and $$b^{2}=\dfrac{4}{3}$$. Substituting these values into the formula gives

$$e = \sqrt{1+\frac{\dfrac{4}{3}}{4}}.$$

Inside the square root, the fraction simplifies as follows:

$$\frac{\dfrac{4}{3}}{4} = \frac{4}{3}\times\frac{1}{4} = \frac{1}{3}.$$

Hence the expression under the radical becomes

$$1 + \frac{1}{3} = \frac{4}{3}.$$

Therefore, we obtain

$$e = \sqrt{\frac{4}{3}}.$$

Taking the square root of the numerator and the denominator separately, we arrive at

$$e = \frac{2}{\sqrt{3}}.$$

Thus the eccentricity of the hyperbola is $$\dfrac{2}{\sqrt{3}}$$.

Hence, the correct answer is Option C.