The artistic crop isn't helpful; it cuts off some vertex names.
The circumcenter H is the meet of the perpendicular bisectors of the sides, helpfully drawn. We have right triangle ELH, right angle L, so
EH² = HL² + EL²
EL = √(EH² - HL²) = √(5.06²-2.74²) ≈ 4.2539393507665339
Since HL is a perpendicular bisector of EF, we get congruent segments FL=EL.
Answer: 4.25
I've drawn a graph in order to a better understanding of this problem. We know that:
BC is perpendicular to AC
∠DBE = 2x - 1
∠CBE = 5x - 42
Let's call the intersection of line BC and AC the point P, so:
∠P=90°
And points B, P and C form the triangle ΔBPC. On the other hand, ∠CBE and ∠PCB are Alternate Interior Angles, so:
∠PCB = ∠CBE = 5x - 42
Moreover:
∠PBC = 2x - 1 - (5x - 42)
∠PBC = 2x - 1 - 5x + 42
∠PBC = -3x + 41
The internal angles of any triangle add up to 180°. Hence for ΔBPC:
90° + ∠PBC + ∠PCB = 180°
90° - 3x + 41 + 5x - 42 = 180°
2x + 89 = 180
2x = 91
x = 45.5°
