(i) The volume of the cylinder is given by \(\pi r^2 h = 250\pi\). Solving for \(h\), we get \(h = \frac{250}{r^2}\).

The total surface area \(S\) is the sum of the lateral surface area and the area of the two bases: \(S = 2\pi rh + 2\pi r^2\).

Substitute \(h = \frac{250}{r^2}\) into the surface area formula:

\(S = 2\pi r \left( \frac{250}{r^2} \right) + 2\pi r^2 = \frac{500\pi}{r} + 2\pi r^2\).

(ii) To find the stationary value, differentiate \(S\) with respect to \(r\):

\(\frac{dS}{dr} = 4\pi r - \frac{500\pi}{r^2}\).

Set \(\frac{dS}{dr} = 0\) to find the critical points:

\(4\pi r = \frac{500\pi}{r^2}\).

Solving for \(r\), we get \(r^3 = 125\), so \(r = 5\).

Substitute \(r = 5\) back into the expression for \(S\):

\(S = 2\pi (5)^2 + \frac{500\pi}{5} = 50\pi + 100\pi = 150\pi\).

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

\(\frac{d^2S}{dr^2} = 4\pi + \frac{1000\pi}{r^3}\).

Since \(\frac{d^2S}{dr^2} > 0\) for \(r = 5\), the stationary value is a minimum.