Solution:

Since $\angle ADC = 90^{\circ},$ we get that $$\angle FDE = 360^{\circ} - 90^{\circ} - 110^{\circ}$$$$= 160^{\circ}.$$ Also since $\triangle FDE$ is isosceles, we get that $$\angle EFD = \dfrac{180^{\circ} - 160^{\circ}}{2} = 10^{\circ}.$$ Finally, we get that $$\angle AFE = 180^{\circ} - 10^{\circ} = 170^{\circ}.$$

Thus, D is the correct answer.