Let $t$ be TNYWR.

The sum of the even integers from 2 to $2 k$ inclusive equals $t$ for some positive integer $k$. That is,

$$
2+4+6+\cdots+(2 k-2)+2 k=t
$$

What is the value of $k$ ?
Manipulating the left side,

$$
\begin{aligned}
2+4+6+\cdots+(2 k-2)+2 k & =t \\
2(1+2+3+\cdots+(k-1)+k) & =t \\
2\left(\frac{1}{2} k(k+1)\right) & =t \\
k(k+1) & =t
\end{aligned}
$$

Since the answer to (a) is 132 , then $t=132$.

Since $k(k+1)=132$ and $k$ is positive, then $k=11$.