To find out how many times the painter will be required to write zero while serially numbering the flats from 1 to 200, we need to consider the numbers from 1 to 200 and count the occurrence of zero.

Let's break down the problem into smaller parts to make it easier to solve:

Counting the occurrence of zero from 1 to 9:
- In this range, there is only one single-digit number that contains zero, which is zero itself.

Counting the occurrence of zero from 10 to 99:
- In this range, there are ten numbers that have zero in the units place: 10, 20, 30, 40, 50, 60, 70, 80, 90.
- Additionally, there are nine numbers that have zero in the tens place: 0, 10, 20, 30, 40, 50, 60, 70, 80. However, we have already counted these numbers in the previous step.
- So, there are a total of 10 numbers in the units place and 9 numbers in the tens place that contain zero.

Counting the occurrence of zero from 100 to 199:
- In this range, there are ten numbers in the units place that contain zero, just like in the previous step.
- Additionally, there are ten numbers in the tens place that contain zero: 100, 101, 102, ..., 109.
- However, we have already counted ten numbers in the units place in the previous step, so we don't need to count them again.
- So, there are a total of 10 numbers in the units place and 10 numbers in the tens place that contain zero.

Counting the occurrence of zero in 200:
- The number 200 contains zero in both the units and tens place.

Now, let's add up the counts from each range:

Count of zero in the units place = 10 + 10 + 1 = 21
Count of zero in the tens place = 9 + 10 = 19

Finally, we add the count of zero in 200, which is 1.

Total count of zero = Count in units place + Count in tens place + Count in 200
= 21 + 19 + 1
= 41

Therefore, the painter will be required to write zero 41 times while serially numbering the flats from 1 to 200.

Since none of the provided answer options match with the correct count of 41, we can conclude that none of the options are correct for this question.

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