r is a positive real number less than 1/1000

Since we want $\sqrt[3]{m}$ to be just a little larger than an integer, $m$ should be just a bit over a perfect cube. Let $m=n^{3}+1$ .

You could use some fancy algebra to find the solution, but 1/1000 isn't really all that small and the answer shows up quickly with a table of $n$ :

$\sqrt[3]{19^{3}+1}=\boxed{19}.00092332$

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Edit: ok let's see how the fancy algebra works in case someone decides to redo this with a really small value of $r$ .

We seek to make $\sqrt[3]{n^{3}+1}-\sqrt[3]{n^{3}}<0.001$

$\sqrt[3]{n^{3}+1}<n+0.001$ cube both sides is ok since $n>1$

$n^{3}+1<n^3+3n^{2}\cdot0.001+3n\cdot0.001^{2}+0.001^{3}$

$0.003n^{2}+0.000003n-0.999999999>0$ and applying the quadratic formula gives

$n>18.2569$ or $n\ge \boxed{19}$