jmc

We compute the powers of 5 modulo 1000: $$\begin{array}{rl}{5}^0& \equiv 1(\mathrm{mod}1000)\\ 5^1& \equiv 5(\mathrm{mod}1000)\\ 5^2& \equiv 25(\mathrm{mod}1000)\\ 5^3& \equiv 125(\mathrm{mod}1000)\\ 5^4& \equiv 625(\mathrm{mod}1000)\\ 5^5& \equiv 125(\mathrm{mod}1000).\end{array}$$ This pattern repeats every two terms starting at the 4th term. In particular, when $n>2$, and $n$ is odd, $$5^n\equiv 125(\mathrm{mod}1000).$$ Therefore the rightmost digit of $5^{1993}$ is $\boxed{125}$.