The surface charge density of a thin charged disc of radius R is $$\sigma$$. The value of the electric field at the centre of the disc is $$\frac{\sigma}{2\epsilon_0}$$. With respect to the field at the centre, the electric field along the axis at a distance R from the centre of the disc:

The electric field at a point along the axis of a uniformly charged disc is given by the formula:

$$E(x) = \frac{\sigma}{2\epsilon_0} \left(1 - \frac{x}{\sqrt{x^2 + R^2}}\right)$$

At the center of the disc, where $$x = 0$$, the electric field is:

$$E(0) = \frac{\sigma}{2\epsilon_0} \left(1 - \frac{0}{\sqrt{0^2 + R^2}}\right) = \frac{\sigma}{2\epsilon_0} (1 - 0) = \frac{\sigma}{2\epsilon_0}$$

This matches the given value. Now, we need to find the electric field at a distance $$R$$ from the center along the axis, so we substitute $$x = R$$:

$$E(R) = \frac{\sigma}{2\epsilon_0} \left(1 - \frac{R}{\sqrt{R^2 + R^2}}\right)$$

Simplify the denominator:

$$\sqrt{R^2 + R^2} = \sqrt{2R^2} = R\sqrt{2}$$

So,

$$E(R) = \frac{\sigma}{2\epsilon_0} \left(1 - \frac{R}{R\sqrt{2}}\right) = \frac{\sigma}{2\epsilon_0} \left(1 - \frac{1}{\sqrt{2}}\right)$$

Let $$E_c = E(0) = \frac{\sigma}{2\epsilon_0}$$ denote the field at the center. The reduction in the electric field at $$x = R$$ compared to the center is $$E_c - E(R)$$. The percentage reduction with respect to $$E_c$$ is:

$$\text{Percentage reduction} = \frac{E_c - E(R)}{E_c} \times 100\%$$

Substitute the expressions:

$$\frac{E_c - E(R)}{E_c} = \frac{\frac{\sigma}{2\epsilon_0} - \frac{\sigma}{2\epsilon_0} \left(1 - \frac{1}{\sqrt{2}}\right)}{\frac{\sigma}{2\epsilon_0}}$$

The common factor $$\frac{\sigma}{2\epsilon_0}$$ cancels out:

$$= \frac{1 - \left(1 - \frac{1}{\sqrt{2}}\right)}{1} = \frac{1 - 1 + \frac{1}{\sqrt{2}}}{1} = \frac{1}{\sqrt{2}}$$

Therefore,

$$\text{Percentage reduction} = \frac{1}{\sqrt{2}} \times 100\%$$

Calculate the numerical value:

$$\frac{1}{\sqrt{2}} \approx 0.7071$$

$$0.7071 \times 100\% = 70.71\%$$

Rounding to one decimal place, this is approximately 70.7%.

Hence, the electric field reduces by 70.7% with respect to the field at the center.

So, the answer is Option A.