(i) The volume of the prism is given by \(\frac{1}{2} \times \frac{\sqrt{3}}{2} x^2 \times h = 2000\). Solving for \(h\), we have:

\(h = \frac{8000}{\sqrt{3}x^2}.\)

The total surface area \(A\) is the sum of the areas of the three rectangles and the two triangular bases:

\(A = 3xh + 2 \times \frac{1}{2} \times \frac{\sqrt{3}}{2} x^2.\)

Substituting for \(h\), we get:

\(A = \frac{\sqrt{3}}{2}x^2 + \frac{24000}{\sqrt{3}}x^{-1}.\)

(ii) To find the stationary value, differentiate \(A\) with respect to \(x\):

\(\frac{dA}{dx} = \frac{\sqrt{3}}{2} \times 2x - \frac{24000}{\sqrt{3}}x^{-2}.\)

Set \(\frac{dA}{dx} = 0\):

\(\sqrt{3}x - \frac{24000}{\sqrt{3}}x^{-2} = 0.\)

Solving for \(x\), we find:

\(x^3 = 8000 \Rightarrow x = 20.\)

(iii) To determine the nature of the stationary value, find the second derivative:

\(\frac{d^2A}{dx^2} = \sqrt{3} + \frac{48000}{\sqrt{3}}x^{-3}.\)

Since \(\frac{d^2A}{dx^2} > 0\), the stationary value is a minimum.