resistance of galvanometer is 17 Ω
For loop ABDA:
55IAB+17IG−21IBD=0
For loop BCDB:
25(IAB−IG)−45(IBD+IG)−17IG=0
For loop ADCEA:
21IBD+45(IBD+IG)=24
So, we have:
55IAB+17IG−21IBD=0
25IAB−87IG−45IBD=0
66IBD+45IG=24⟹22IBD+15IG=8
IBD=0.36−0.68IG
55IAB+31.28IG=7.56
25IAB−56.4IG=16.2
IAB=0.65+2.26IG
35.75+124.3IG+31.28IG=7.56
the current through the galvanometer:
I_G=(7.56-35.75)/(124.3+31.28)=-0.18\A
the effective resistance of the network:
R1=171+45+251+55+211==904405320+1292+1190=904407802
R=780290440=11.6 Ω
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