jmc

To find the tens digit of $17^{1993}$, we can look at the first few power of 17 modulo 100: $$\begin{array}{rl}17^0& \equiv 1,\\ 17^1& \equiv 17,\\ 17^2& \equiv 17\cdot 17\equiv 289\equiv 89,\\ 17^3& \equiv 17\cdot 89\equiv 1513\equiv 13,\\ 17^4& \equiv 17\cdot 13\equiv 221\equiv 21(\mathrm{mod}100).\end{array}$$

We know that if we find a power of 17 whose last two digits are 01, then the last two digits in the power of 17 become periodic at that point. We don't have that in $17^4$, but the units digit in $17^4$ is 1. We have matched the units digit, so let's use powers of $17^4$: $$\begin{array}{rl}17^4& \equiv 21,\\ 17^8& \equiv 21\cdot 21\equiv 441\equiv 41,\\ 17^{12}& \equiv 21\cdot 41\equiv 861\equiv 61,\\ 17^{16}& \equiv 21\cdot 61\equiv 1281\equiv 81,\\ 17^{20}& \equiv 21\cdot 81\equiv 1701\equiv 1(\mathrm{mod}100).\end{array}$$ We found a power of 17 whose last two digits are 01, so the last two digits are periodic, with period 20.

Since $1993\equiv 13(\mathrm{mod}20)$, $$17^{1993}\equiv 17^{13}(\mathrm{mod}100).$$ Then $$\begin{array}{rl}17^{13}& \equiv 17^{12}\cdot 17\\ & \equiv 61\cdot 17\\ & \equiv 1037\\ & \equiv 37(\mathrm{mod}100).\end{array}$$ Therefore, the tens digit of $17^{1993}$ is $\boxed{3}$.