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Let's write the lengths of the 3 pieces : a, b, and c.
We know that a+b+c = 2/3 m = 1,5 m = 150 cm

(1) one length's piece is 20 cm

for example a= 20 cm
but b and c can be every couple such as b+c = 150 - 20 = 130 cm (100,30 or 70,60 or 50,80...)

Not sufficient

(2) The sum of the lengths of two of the pieces is equal to the length of the third piece

for example a+b = c
and a+b+c = 150
so 2c=150 => c = 75
but a and b can be every couple such as a+b = 150-75 = 75 (70,5 or 60,15 ...)

Not sufficient

(combined)

c=75
a=20
so, b=150-75-20 = 55

Sufficient, answer choice C
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Say each piece be a,b,c cms.

Given : a+b+b = 150 cms

From

1) a or b or c is 20 cms. No info about rest 2. Insufficient.

2) say a+b = c => c =75. No info about and b. Insufficient.

Combining 1 and 2

c = 75 , say a= 20 => b=55. Hence 3 lengths are 20,55,75. Sufficient,

Hence C
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Remember we don't need to get the exact answers in DS questions.

From info available l1+l2+l3 = 150 cms and we need to get l1, l2 and l3.

S1: l1 = 20 Not sufficient.

S2: l1 + l2 = l3.
So, 2 l3 = 150 or l3 = 75. So l1+l2 = 75 . Not sufficient to get l1 & l2.

S1 & S2 we can get l1 = 20 & l1 + l2 = 75, so C is sufficient to get the answer.
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