#### Answer

(e) is the correct answer.

#### Work Step by Step

Since the block is suspended and has no acceleration, if the block has mass $m$ and the tension force in the rope is $T$, we have $$\sum T=mg$$
The tension force $T$ is unchanged in a specific rope, but the way the rope is arranged can affect the number of $T$ forces affecting the block.
In A, there is only 1 section of rope holding the block, so $\sum T=T_A$. Therefore, $T_A=mg$
In B, the rope is divided among 3 pulleys, which separate the rope into 3 parts all helping to keep the block in place. Each section of the rope has tension $T$, so $\sum T=3T_B$. Therefore, $T_B=\frac{mg}{3}$
In C, the rope is divided among 2 pulleys, which separate the rope into 2 parts. Each part has tension $T$, so $\sum T=2T_C$. Therefore, $T_C=\frac{mg}{2}$
In conclusion, the tension in ascending order is: B, C, A.