jmc

Since we want the tens digit, we can find the last two digits of $7^{2011}$. We can do this by using modular arithmetic. $$7^1\equiv 07(\mathrm{mod}100).$$ $$7^2\equiv 49(\mathrm{mod}100).$$ $$7^3\equiv 43(\mathrm{mod}100).$$ $$7^4\equiv 01(\mathrm{mod}100).$$ We can write $7^{2011}$ as $(7^4{)}^{502}\times 7^3$. Using this, we can say: $$7^{2011}\equiv (7^4{)}^{502}\times 7^3\equiv 7^3\equiv 343\equiv 43(\mathrm{mod}100).$$ From the above, we can conclude that the last two digits of $7^{2011}$ are 43. Since they have asked us to find the tens digit, our answer is $\boxed{4}$.