The final expression is in terms of the length of a median and the area of the triangle. Translations will not affect those values. So then we can translate the parabola so its vertex is at the origin: y=ax^2.
Let point A be ( p,ap^2 ) and point B be ( q,aq^2 ). Then the midpoint M (aka C1 in the problem statement) is ( (p+q)/2,(a/2)*(p^2+q^2) ).
The slope of the tangent at point A is 2ap, then the equation of that line is y = 2apx - ap^2.
The slope of the tangent at point B is 2aq, then the equation of that line is y = 2aqx - aq^2.
The intersection point C is then ( (p+q)/2,apq ).
C and M have the same x-coordinate.
Then the length CM is simply (a/2)*(p^2+q^2) - apq = (a/2)*(p-q)^2.
The area of triangle ABC can be found by using the determinant formula with three coordinate points. When reduced, the area equals (a/4)*abs( (p-q)^3 ).
So then the final ratio we are to compute is Area^2/CM^3
= [(a/4)*abs( (p-q)^3 )]^2 / [ (a/2)*(p-q)^2 ]^3
= (a^2/16) / (a^3/8) = 1/(2a).
Solution
