Since we want the tens digit, we can find the last two digits of $7^{2011}$. We can do this by using modular arithmetic.  $$ \begin{aligned} 7 \equiv 07 \&(\bmod 100) \textbackslash\textbackslash{} 7^{2} \equiv 49 \&(\bmod 100) \textbackslash\textbackslash{} 7^{3} \equiv 43 \&(\bmod 100) \textbackslash\textbackslash{} 7^{4} \equiv 01 \&(\bmod 100) \end{aligned} $$  We can write $7^{2011}$ as $\left(7^{4}\right)^{502} \times 7^{3}$. Using this, we can say: $$ 7^{2011} \equiv\left(7^{4}\right)^{502} \times 7^{3} \equiv 7^{3} \equiv 343 \equiv 43 \quad(\bmod 100) . $$  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 (D) $4$ 
What is the tens digit of $7^{2011}$? (2011 AMC 8 Problems, Question \#22) 
[A] $$0$$ 
[B] $$1$$ 
[C] $$3$$ 
[D] $$4$$ 
[E] $$7$$ 