Prakash should not test if the number of bad widgets in the lot is:

We should consider three possible cases as 1. when he is using test 1
2. When he is using test 2
3. When he is using no test
Now we will choose a method where the total expenditure will be least.
So for option A, let's consider that number of defective pieces are 50.
Hence, while using test 1:
Cost of testing = $$1000 \times 2 = 2000$$
Correcting 80% of pieces = $$40 \times 25 = 1000$$
Penalty for 20% of pieces = $$10 \times 50 = 500$$
Total = 3500

For test 2:
Cost of testing = $$1000 \times 3 = 3000$$
Correcting all 50 pieces = $$ 50 \times 25 = 1250 $$
Total= 4250

For no test:
Penalty for all 50 pieces = $$ 50 \times 50 = 2500$$

Hence cost is least when he is using no test while number of defective pieces is less than 100.