, 2003

Solution:

The answer is (D). If we denote by $x, y$ and $z$ the lengths of the sides in increasing order, the hypotheses translate as
$$\left\{ \begin{array}{l} x y + x z + y z = \frac{228}{2} = 114 \\ x^{2} + y^{2} + z^{2} = 133 \\ x z = y^{2} \end{array} \right.$$

From the first two we obtain $(x + y + z)^{2} = \left(x^{2} + y^{2} + z^{2}\right) + 2(x y + x z + y z) = 361$ and hence $x + y + z = 19$. Substituting the third into the first we obtain $y(x + y + z) = x y + x z + y z = 114$, from which $y = 6$. Therefore the volume is $x y z = y^{3} = 216$. On the other hand, a parallelepiped with sides $x = 4, y = 6, z = 9$ satisfies the hypotheses.