Explanation

This is a geometric-progression rebound problem (the options print mixed fractions, e.g. 102/3 means 10 2/3). Each bounce reaches 2/3 of the previous height, so after n bounces the height is 36 × (2/3)ⁿ. For the third bounce: 36 × 8/27 = 288/27 = 32/3 = 10⅔ m, i.e. option (b). The trap is stopping at two bounces (16 m) or forgetting the initial 36 m factor. Hence option (b).