Answer
The tension in the cable is $3125~N$
Work Step by Step
When the cable sags, both ends of the cable make an angle of $\theta$ below the horizontal. Note that $sin~\theta = \frac{0.12~m}{3.00~m}$
We can consider the left side and the right side of the cable. The sum of the vertical component of the tension $T$ in each side of the cable is equal in magnitude to Denisha's weight.
We can find the tension $T$ in the cable:
$2T~sin~\theta = 250~N$
$T = \frac{250~N}{2~sin~\theta}$
$T = \frac{250~N}{(2)~(\frac{0.12~m}{3.00~m})}$
$T = 3125~N$
The tension in the cable is $3125~N$.
