method 1. using method of joints

normally, 1st step is to work out support reactions at J & K. however, since the question only requires force in member IK, so not necessary to find support reactions.

by considering vertical & horizontal forces at each joint, progressively work way through the joints one-by-one in following order ...
A : gives force in members AD(0), AB(-100)
C : gives force in members CF(0), BC(0)
B : gives force in members BD(80.266), BF(-80.266)
D : gives force in members DG(65.489), DE(-131.395)
F : gives force in members FI(-65.489), EF(31.395)
E : gives force in members EG(110.341), EI(-110.341)
I : gives force in member IK(-143.188)
[this method is laborious.
maintain more significant figures in intermediate values in order to have better accuracy in final answer.]

method 2. using method of sections

consider a cut through members DG, EG, IH & IK, splitting the truss into 2 sections.

consider just the upper section as a free body. the external forces acting on the upper section are ...
the 100 kN forces at A & D, and ...
the forces on cut members DG,EG,IH & IK.

taking moments about point G, the forces in members DG, EG & IH do not contribute any moments because they all pass through G.
therefore, sum of moments
100 x 2.7 + 100 x 5.4 + (force in IK) x GI sin GIF = 0
810 + (force in IK) x 5.9 sin73.496 = 0
gives force in IK = -143.188
[much lesser calculations than method 1.]