Let the Four consecutive odd integers is 2n-3,2n-1,2n+1,2+3

And thier average= 42

and thier sum is = 42×4= 168

⇒  (2n-3)(2n-1)+(2n+1)+(2+3)=168

⇒  8n=168

⇒  n=21

∴ The numbers are J=39, K=41, L=43, M=45

The product of K and M is 41×45=1845.

Hence, Option D is correct.