Solution:

From the first equation $x^{2} y = 150$ we obtain $y = \dfrac{150}{x^{2}}$ (with $x \neq 0$).

We substitute into the second equation:
$$x^{3} y^{2} = 4500$$
Substituting $y$:
$$x^{3} \left( \dfrac{150}{x^{2}} \right)^{2} = 4500$$
$$x^{3} \cdot \dfrac{22500}{x^{4}} = 4500$$
$$\dfrac{22500 x^{3}}{x^{4}} = 4500$$
$$\dfrac{22500}{x} = 4500$$
$$x = \dfrac{22500}{4500} = 5$$

Now we compute $y$:
$$y = \dfrac{150}{x^{2}} = \dfrac{150}{25} = 6$$

Let us check whether there are other solutions:

We assumed $x \neq 0$. Let us also consider $x < 0$:
$$x = -5$$
$$y = \dfrac{150}{(-5)^{2}} = \dfrac{150}{25} = 6$$

Let us check whether the pair $(x, y) = (-5, 6)$ satisfies the second equation:
$$x^{3} y^{2} = (-5)^{3} \cdot 6^{2} = (-125) \cdot 36 = -4500$$
This is not equal to $4500$, so it is not a solution.

Let us consider $y < 0$:
From the first equation $x^{2} y = 150$, if $y < 0$ then $x^{2} < 0$, which is impossible.

Therefore the only real solution is $(x, y) = (5, 6)$.

Correct answer: (B) one.