From the question stem, we are able to get the following details

1. Each sealed packet has a marker - Red, Blue, Or Black(which we can't see from outside)

2. 23 of the 40 packets have a black marker. Therefore, Blue + Red = 17

1. Statement 1 tells us that Blue < Red

Case 1: Red = 13, Blue = 4

Case 2: Red = 10, Blue = 7

Case 3: Red = 9, Blue = 8

We cannot come to a unique value for the number of packets containing Blue markers

(Insufficient)2. If he needs to withdraw 20 markers to have 8 of a color,

2 cases are possible such that 8 markers of either color can be taken out.

Case 1: Red = 12, Blue = 5

Case 2: Red = 5, Blue = 12

We cannot come to a unique value for the number of packets containing Blue markers

(Insufficient)Combining the information from both the statements,

the only option possible is Black = 23, Blue = 5, Red = 12

(Sufficient - Option C)
_________________

You've got what it takes, but it will take everything you've got