(i) The volume of the block is given by \(V = 2x \cdot x \cdot y = 72\). Thus, \(2x^2y = 72\) which implies \(y = \frac{72}{2x^2} = \frac{36}{x^2}\).

The total surface area \(A\) is given by:

\(A = 2(2x \cdot x + x \cdot y + 2x \cdot y) = 2(2x^2 + xy + 2xy) = 4x^2 + 6xy\).

Substitute \(y = \frac{36}{x^2}\) into the equation:

\(A = 4x^2 + 6x \cdot \frac{36}{x^2} = 4x^2 + \frac{216}{x}\).

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

\(\frac{dA}{dx} = 8x - \frac{216}{x^2}\).

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

\(8x - \frac{216}{x^2} = 0\).

\(8x^3 = 216\) which gives \(x^3 = 27\) and \(x = 3\).

(iii) Substitute \(x = 3\) back into the expression for \(A\):

\(A = 4(3)^2 + \frac{216}{3} = 36 + 72 = 108\) cm².

To determine if this is a minimum or maximum, find the second derivative:

\(\frac{d^2A}{dx^2} = 8 + \frac{432}{x^3}\).

At \(x = 3\), \(\frac{d^2A}{dx^2} = 8 + \frac{432}{27} = 24\), which is positive, indicating a minimum.