You are right! The sample size of the two groups are very different, the cultivated group with 157 individuals, while the wild group is only 14 individuals.
What do you mean by comparing the distance R^2=0.2? Just by saying:
Genome-wide linkage disequilibrium (LD) analyses revealed a faster LD decay in wild (r2 ∼ 0.2 at 400 bp) compared to cultivated group(r2 ∼ 0.20 at 70 kb)
Reviewers asked for a statistical test to support this sentence (>_<")

Something looks weird in the blue line, it looks like it's not decaying, can it be that every pair of SNPs will have a correlation r^2 of 0.15? Maybe you have a bias in the population? I'm not sure if there is a test to compare the two plots but you can take the same pairs of SNPs and compute the correlation between them in both populations then compare the coefficients using a paired t-test.
This was my initial plan, but using paired t-test you compare the means between two related groups. Does it make sense to compare the two means of curves?
No, that's not what I meant. Select a group of SNP pairs with a certain distance range (100kb +- 10kb for instance) then for each pair compute the correlation using each population, you now have paired measurements that you can compare. You can repeat the process for several point along the curve. The LD drop curve is deceiving, it's a smoothing of averaging of r^2 values in bins of distances, it looks pretty but actually very much noisy.
Very good. Now I understand your point. Thanks a lot!
And with vcftools http://vcftools.sourceforge.net/man_latest.html I can define the physical window
Example:
--ld-window-bp 100000compute the LD between SNPs being maximum 100 kb apart