3 cooks have to make 80 burgers.They are known to make 20

The question looks calculation intensive but it isn't. You only have to use some logical hit and trial.

We know that the three together cook 20 burgers in 1 min i.e. we have their combined rate:
ra + rb + rc = 20 burgers/min

Also, they made 80 burgers in 8 mins such that:
\(\frac{20}{ra} + \frac{60}{(rb + rc)} = 8\)

Now, what is the significance of 'The 1st cook began workin alone and made 20 pieces having worked for sometime more than 3 mins'
We can't work with 'something more than 3 mins' but it does tell us something. All three together take 1 min to make 20 burgers but cook 1 takes more than 3 mins to make 20 burgers so cook 1 is slower than the average rate. If all three had the same rate, all three would have taken 3 mins each to cook 20 burgers.

Let's assume a value for ra. We know it is less than the average.
ra + rb + rc = 20 burgers/min
The combined rate is 20 so average rate would be 6.666. Let's assume ra = 5 and rb+rc = 15 since these values are appropriate for the second equation too.

\(\frac{20}{ra} + \frac{60}{(rb + rc)} = 8\)
\(\frac{20}{5} + \frac{60}{(15)} = 8\)

This does hold! This means ra = 5 burgers/min.
So time taken to make 160 burgers = 160/5 = 32 mins

Note that had ra= 5 not worked, I would have tried ra = 4 since it gives simple values too