Data Given: $Volume \ of \ sample = 25\ c.c, \ \ \ Volume \ of \ K_2Cr_2O_7: \ 8.3\ c.c, \ \ \ Molarity \ of \ K_2Cr_2O_7:\ \ \ 0.001\ M$

$Since \ 1\ mole\ FAS \equiv 1\ mole\ K_2Cr_2O_7\equiv \dfrac{1}{2}mole\ of \ O_2$

$Hence, \ 8.3\ ml\ K_2Cr_2O_7\equiv same \ Volume \ of \ FAS\ assuming\ same\ molarity\\ Hence, if\ 1\ litre \ of \ 1M\ FAS=8\ g \ of \ oxygen\\ 8.3\ ml \ of \ 0.001M\ FAS=\dfrac{8\times 8.3\times 0.001}{1000}=7.04\times 10^{-5}\ g\ of\ oxygen\ for\ 25\ ml\ of\ sample$

Oxygen required for 1000 ml o waste water $=7.4\times 10^{-5}\times 4=2816\times 10^{-3}\ g/lit$

COD for waste water sample = $2.816\times 10^{-3}\ mg/lit$