I think it’s 50.
The inscribed angle theorem states that an angle θ inscribed in a circle is half of the central angle 2θ that subtends the same arc on the circle, like the the picture at the end shows.
So the measure of ∠COB should twice of the measure of ∠CDB, and thus ∠COB=130°
Then as the problem said “Angle A is circumscribed about circle O “,both ∠ACO and ∠ABO should be right angle.
Finally, using the property that “the sum of the quadrangle’s interior angles is 360°”, we can conclude that
∠CAB=360°-∠COB-∠ACO-∠ABO=360°-130°-90°-90°=50°
