Solution:

Let $\theta$ be the measure of $\angle BOC.$

One component of the shaded region is the area of the circle with radius $2$ minus the area of the circle with radius $1.$ This part has area $4\pi-\pi=3\pi.$ The remaining area is a sector of the biggest circle minus the area of the circle with radius $2$. This has area $\dfrac{\theta}{360}(9\pi-4\pi) = \dfrac{\theta}{360}(5\pi).$ Hence, the total area of the shaded region is $3\pi + \dfrac{\theta}{360}(5\pi).$

Next, we note that the unshaded region is composed of the smallest circle and the unshaded portion of the outer ring. This will have a total area of $\pi + \dfrac{360-\theta}{360}(5\pi)$

Lastly, we equate the area of both regions and solve for $\theta:$ $$3\pi + \dfrac{\theta}{360}(5\pi) = \pi + \dfrac{360-\theta}{360}(5\pi)$$ $$2\pi = \dfrac{360-\theta-\theta}{360}(5\pi)$$ $$\dfrac{2}{5} = 1 - \dfrac{2\theta}{360}$$ $$2\theta = \dfrac{3}{5}(360)$$ $$\theta = 108.$$

Thus, A is the correct answer.