$121_b$ can be represented in base 10 as $b^2+2b+1$, which factors as $(b+1)^2$.
Note that $b>2$ because 2 is a digit in the base-b representation, but for any 
$b>2$, $121_b$ is the square of $b+1$. $\boxed{\textbf{(D)}}$
The number $121_b$, written in the integral base $b$, is the square of an integer, for 
$\textbf{(A)}\ b = 10,\text{ only}\qquad\textbf{(B)}\ b = 10\text{ and }b = 5,\text{ only}\qquad$
$\textbf{(C)}\ 2\leq b\leq 10\qquad\textbf{(D)}\ b > 2\qquad\textbf{(E)}\ \text{no value of }b$