jmc

Since $y$ and $\sqrt{x}$ are inversely proportional, this means that $y\sqrt{x}=k$ for some constant $k$. Substituting the given values, when $x=24$ and $y=15$, we find that $15\sqrt{24}=30\sqrt{6}=k$. Therefore, when $y=3$, we can solve for $x$: $$\begin{array}{rl}3\cdot \sqrt{x}& =30\sqrt{6}\\ \Rightarrow (\sqrt{x}{)}^2& =(10\sqrt{6}{)}^2\\ \Rightarrow x& =100\cdot 6\\ & =\boxed{600}\end{array}$$