Total pastries = 400
Let pastries with chocolate only = a
Let pastries with nuts only = b
Let pastries with both chocolate and nuts = c
Let pastries without chocolate and nuts = d
400 = a + b + c + d

Given that 60% contained chocolate; then c = 60% of 400 = 240

Target question => How many of those sold contained only nuts? i.e find b

b + d + 240 = 400 and b = 400 - 240 - d
b = 160 - d

Statement 1 => The number of pastries containing neither is one-fourth of the number containing chocolate
i.e d = 1/4 of (a + c) = 1/4 * 240 = 60
Fron question stem, b = 160 - 60 = 100
Statement 1 is SUFFICIENT

Statement 2 => One-third of pastries sold containing chocolate also contained nuts
$$i.e\ c=\frac{1}{3}of\ 400\approx133$$
This cannot give the estimate of b because d is still unknown

Statement 1 alone is SUFFICIENT
Answer = A