A sample of pond water containing 20mg of organic matter requires 16mg of dissolved oxygen. (Pond water contains 10mg of organic matter per 2 liters). Its BOD is

 It is given that the pond water contains 10mg of organic matter per 2 liters. This means that for 1mg of organic matter it will be 0.2L. Thus, for 20mg of organic matter, it will be 4L.

Now,

$B.O.D=\frac{no.of mg oxygen required}{volumeofwater}=\frac{16}{4}mg /mL=4 ppm$

Thus, the $BOD$ will be $4 ppm$.