jmc

Begin by factoring $24$ and $48$. We have $24=2^3\cdot 3$ and $48=2^4\cdot 3$, so $$7\cdot 24\cdot 48=7\cdot (2^3\cdot 3)\cdot (2^4\cdot 3)=2^7\cdot 3^2\cdot 7.$$For a number to be a perfect cube, every prime factor must have an exponent that is a multiple of $3$. The next multiple of $3$ bigger than $7$ is $9$, so we need a $2^2$ to reach $9$ in the exponent. We need one more factor of $3$ to reach $3^3$. We need another $7^2$ to reach $3$ in the exponent of $7$. This gives a smallest number of $2^2\cdot 3\cdot 7^2=\boxed{588}$.