Another confusion......Arrange them in increasing order of stability.......

1>2, and 3>4
but not sure how 1 and 3 compare

I don't see how the 3rd radical is any different from the 4th. The radical doesn't delocalise over benzene, try making the structures and see for yourself! Had benzene been attached to the double bonded carbon, the 3rd radical would be the clear winner in this little competition.

I would rather say 3 = 4 > 1 > 2.

@pritish, doesnt phenyl grp have a weak -I effect?

so, maybe it would be 4>~3>1>2

well the answer should be 4≈3>1>2 :

REASONS:
1.inductive effect dies out after three bonds!
2.inudctive effect due to phenyl grp is weak -I effect!

REASON FOR 1>2:
1.we see tht 1 has a -CH₃ group too which adds on to the stability of 1...so 1>2

The ans is 3>4>1>2

Possibly so asish, but Ph's -I effect is very weak. I would still go with 3 = 4. Euracle that answer is not possible. Had Ph been attached to the double bonded carbon instead, it would've been the answer you mentioned.

wat are u guys talking here about ?

why are u guys even considering -I effect on a species wich is neutral ?

i think i will go with 4 > 3 > 1 > 2

bcz HC structures in 4 > 3 ....

in no 3

for ph group no of resonance structures is 3+1 =4
so its more stabilised

therefore i would go with

3>4>2>1

kalyan can u please show me how is the radical resonating with the ring ?

i think it is 4>3>1>2

3>4>1>2