Solution:

Let $x$ be the side of the base and $y$ the height of a parallelepiped satisfying the given conditions. Then $x$ and $y$ are positive integers and
$$x^2 y = 270000 = 3^3 \cdot 2^4 \cdot 5^4.$$
By the unique factorization of integers, we have $x = 3^a 2^b 5^c$, $y = 3^d 2^e 5^f$, where the exponents satisfy the following conditions:
$$\begin{array}{ll} 2a + d = 3, & 0 \leq a, d \leq 3, \end{array}$$
Thus the only possible cases are:
$$\begin{array}{r} (a, d) = (0,3),\ (1,1) \\ (b, e) = (0,4),\ (1,2),\ (2,0) \\ (c, f) = (0,4),\ (1,2),\ (2,0) \end{array}$$
Combining in all possible ways the two possibilities for $(a, d)$, the three possibilities for $(b, e)$, and the three possibilities for $(c, f)$, we obtain $2 \cdot 3 \cdot 3 = 18$ parallelepipeds.