This one is a bit more challenging than the usual fare.
I'm always curious when something's crossed out.
I think you got the hard part right, that the equal arcs imply this is an isosceles triangle, JK=FJ. The angle x we seek is the opposite the base.
The trick is to draw the center C. Then CJ bisects angle x because of the isosceles triangle thing.
We get isosceles triangle FCJ where CF=CJ because they're both radii.
So angle CFJ = angle CJF = x/2.
A radius and a tangent make a right angle, so CFG is right and CFJ is complementary to GFJ=56 degrees.
CFJ = 90 - 56 = 34 degrees
x = 2 CFJ = 68 degrees
Answer: C
