jmc

Three-inch cubes can fill a rectangular box only if the edge lengths of the box are all integer multiples of 3 inches. The largest such box whose dimensions are less than or equal to those of the $6^{\prime \prime }\times 5^{\prime \prime }\times 10^{\prime \prime }$ box is a $6^{\prime \prime }\times 3^{\prime \prime }\times 9^{\prime \prime }$ box. The ratio of the volumes of these two boxes is $$\frac{6\cdot 3\cdot 9}{6\cdot 5\cdot 10}=\frac{3\cdot 9}{5\cdot 10}=\frac{27}{50},$$ which is $\boxed{54}$ percent.