Answer
The area is shrinking at $250 \frac{\text{m}^2}{\text{min}}$.
Work Step by Step
When the boon is being pulled in by 5 m/min., the circumference of the containment area is decreasing by 5 m/min. Thus:
$C = 2\pi r$
$\frac{dC}{dt} = 2\pi \frac{dr}{dt}$
$-5 = 2\pi \frac{dr}{dt}$
$\frac{dr}{dt} = -\frac{5}{2\pi}$.
Using the area formula, $A = \pi r^2$; we solve for $\frac{dA}{
dt}$:
Note that because diameter is 100 m, $r$ is 50 m. We also use $\frac{dr}{dt} = -\frac{5}{2\pi}$ from above.
$A=\pi r^2$
$\frac{dA}{dt} = \pi * 2 * r * \frac{dr}{dt}$
$\frac{dA}{dt} = \pi * 2 * 50 * -\frac{5}{2\pi}$
$\frac{dA}{dt}=-250 \frac{\text{m}^2}{\text{min}}$
