An electric current is flowing through a circular coil of radius R. The ratio of the magnetic field at the centre of the coil and that at a distance $$2\sqrt{2}R$$ from the centre of the coil and on its axis is :

The magnetic field at a point on the axis of a circular coil of radius $$R$$ carrying current $$I$$ is given by the formula:

$$ B = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}} $$

where $$x$$ is the distance from the center of the coil along the axis.

First, we find the magnetic field at the center of the coil. At the center, $$x = 0$$. Substituting $$x = 0$$ into the formula:

$$ B_{\text{center}} = \frac{\mu_0 I R^2}{2(R^2 + 0^2)^{3/2}} = \frac{\mu_0 I R^2}{2(R^2)^{3/2}} $$

Since $$(R^2)^{3/2} = R^{3}$$, we simplify:

$$ B_{\text{center}} = \frac{\mu_0 I R^2}{2 R^3} = \frac{\mu_0 I}{2 R} $$

Next, we find the magnetic field at a distance $$x = 2\sqrt{2}R$$ from the center on the axis. Substituting $$x = 2\sqrt{2}R$$ into the formula:

$$ B_{\text{axis}} = \frac{\mu_0 I R^2}{2(R^2 + (2\sqrt{2}R)^2)^{3/2}} $$

First, compute $$(2\sqrt{2}R)^2$$:

$$ (2\sqrt{2}R)^2 = (2\sqrt{2})^2 \times R^2 = 4 \times 2 \times R^2 = 8R^2 $$

Then, $$R^2 + (2\sqrt{2}R)^2 = R^2 + 8R^2 = 9R^2$$. So,

$$ B_{\text{axis}} = \frac{\mu_0 I R^2}{2(9R^2)^{3/2}} $$

Now, compute $$(9R^2)^{3/2}$$:

$$ (9R^2)^{3/2} = (9)^{3/2} \times (R^2)^{3/2} = (3^2)^{3/2} \times R^{3} = 3^{3} \times R^{3} = 27R^3 $$

Substituting back:

$$ B_{\text{axis}} = \frac{\mu_0 I R^2}{2 \times 27R^3} = \frac{\mu_0 I R^2}{54 R^3} = \frac{\mu_0 I}{54 R} $$

The ratio of the magnetic field at the center to that at the axis point is:

$$ \frac{B_{\text{center}}}{B_{\text{axis}}} = \frac{\frac{\mu_0 I}{2 R}}{\frac{\mu_0 I}{54 R}} = \frac{\mu_0 I}{2 R} \times \frac{54 R}{\mu_0 I} $$

Canceling $$\mu_0$$, $$I$$, and $$R$$:

$$ \frac{B_{\text{center}}}{B_{\text{axis}}} = \frac{54}{2} = 27 $$

Hence, the ratio is 27. Comparing with the options, 27 corresponds to option B.

So, the answer is $$27$$.