If a regular hexagon is inscribed within a circle of diameter of 12cm (radius 6cm) then, since a hexagon is effectively made up of 6 equilateral triangles, the sides of the hexagon are equal to one another, and are equal to the radius of the circle.
If the sides of the hexagon are length 5cm, the hexagon is inscribed within another circle of diameter 10cm.
So the picture I'm getting is of two concentric circles, centre O, the inner of which circumscribes the hexagon.
The locations of points A, B, C and D are not specified in the question.
The angles of a regular hexagon are fixed regardless of its size, each interior angle being 120 degrees.
The angle ABC=120 degrees suggests that A, B and C are three consecutive vertices of the hexagon.
The point D I see as being located on the larger circle as an extension of OB.
The triangle ABC is isosceles with the equal sides being two 5cm sides of the hexagon.
The height of ABC is half the radius of the inner circle, so its length is 2.5cm.
The difference between the radii (OD-OB) of the two circles is 1cm, so the length of line DX (X is the mid-point of AC) from D to AC is 2.5+1=3.5cm. We know that angle BAC is 30 degrees.
AX=sqrt(AB^2-BX^2)=sqrt(25-6.25)=4.33cm (Pythagoras). The tangent of angle CAD=DX/AX=3.5/4.33=0.81.
Angle CAD=tan^-1(0.81)=38.95 degrees.