Let $t$ be TNYWR.

Three siblings share a box of chocolates that contains $t$ pieces. Sarah eats $\frac{1}{3}$ of the total number of chocolates and Andrew eats $\frac{3}{8}$ of the total number of chocolates. Cecily eats the remaining chocolates in the box. How many more chocolates does Sarah eat than Cecily eats?
Sarah eats $\frac{1}{3} t$ chocolates and Andrew eats $\frac{3}{8} t$ chocolates.

Since Cecily eats the rest of the chocolates, she eats $t-\frac{1}{3} t-\frac{3}{8} t=\frac{24-8-9}{24} t=\frac{7}{24} t$.

Therefore, Sarah eats $\frac{1}{3} t-\frac{7}{24} t=\frac{1}{24} t$ more chocolates than Cecily.

Since the answer to (a) is 144, then $t=144$ and so Sarah eats 6 more chocolates than Cecily.