Increasing order of electron withdrawing power of following functional groups is:
a. $$-\text{CN}$$, b. $$-\text{COOH}$$, c. $$-\text{NO}_2$$, d. $$-\text{I}$$

Electron-withdrawing power is determined mainly by two effects:
(i) the inductive effect (−I), which pulls electron density through σ-bonds, and
(ii) the resonance/mesomeric effect (−M when the group pulls electron density through π-bonds, +M when it donates).

Step 1 Analyse each substituent.

1. $$-\text{I}$$ (iodo) : Iodine is less electronegative than N, O, or Cl, so its −I effect is weak. Moreover, thanks to its lone pairs it can donate by +M into an attached π-system, further reducing the net electron-withdrawing ability. Hence this is the weakest EW group in the list.

2. $$-\text{COOH}$$ (carboxyl) : Oxygen atoms are electronegative, so the group shows a moderate −I effect. Inside the group itself the C=O pulls electron density from the O−H bond, but relative to other strong −I groups, $$-\text{COOH}$$ is only moderately electron-withdrawing.

3. $$-\text{CN}$$ (cyano) : The carbon of the triple bond is sp-hybridised (50 % s-character) and directly attached to a highly electronegative nitrogen; therefore the −I effect is strong. There is no +M donation, so the net electron-withdrawing power is high.

4. $$-\text{NO}_2$$ (nitro) : The nitro group possesses two strongly electronegative oxygens and a positively charged nitrogen in its major resonance form $$\text{O}^{-}\!-\!\text{N}^{+}\!=\!\text{O}$$. It shows both a very powerful −I effect and a very powerful −M effect (withdraws electron density via resonance), making it the strongest electron-withdrawing group among common substituents.

Step 2 Arrange in increasing order of withdrawing power (weakest → strongest):
$$-\text{I} \lt -\text{COOH} \lt -\text{CN} \lt -\text{NO}_2$$

This sequence corresponds to Option C.

Final Answer: Option C which is: $$d \lt b \lt a \lt c$$