The length of rope is equal half the circumference of the silo.
$L_{rope} = 10\pi ~ \text{m}$
The total area for grazing is shown above. It is composed of a semi circle and area bounded by involute of circle, the circle, and the vertical line along the quadrant point on the circle.
For the semi-circle
$A_1 = \frac{1}{2}\pi (L_{rope})^2 = \frac{1}{2}\pi (10\pi)^2 $
$A_1 = 50\pi^3 ~ \text{m}^2$
For the involute of circle
Length of rope wrapped on the silo
$s = 10\theta$
Length of unwrapped/stretched rope
$r = L_{rope} - s = 10\pi - 10\theta$
$r = 10(\pi - \theta)$
$\displaystyle A_2 = \frac{1}{2} \int_{\theta_1}^{\theta_2} r^2 \, d\theta$
$\displaystyle A_2 = \frac{1}{2} \int_{0}^{\pi} 100(\pi - \theta)^2 \, d\theta$
$\displaystyle A_2 = 50 \int_{0}^{\pi} (\pi - \theta)^2 \, d\theta$
$\displaystyle A_2 = -50 \int_{0}^{\pi} (\pi - \theta)^2 \, (-d\theta)$
$A_2 = -50 \left[ \dfrac{(\pi - \theta)^3}{3} \right]_{0}^{\pi/2}$
$A_2 = -\frac{50}{3} \left[ (\pi - \pi)^3 - (\pi - 0)^3 \right]$
$A_2 = -\frac{50}{3} (-\pi^3)$
$A_2 = \frac{50}{3}\pi^3 ~ \text{m}^2$
Total area for grazing
$A = A_1 + 2A_2$
$A = 50\pi^3 + 2(\frac{50}{3}\pi^3)$
$A = \frac{250}{3}\pi^3 = 2\,583.86 ~ \text{m}^2$ answer: D
