Sucrose hydrolyses in acidic medium into glucose and fructose by...

For a first order reaction, the amount of reactant remaining after $n$ half-lives is given by $N_t = \dfrac{N_0}{2^n}$.

Given $t_{1/2} = 3$ hours and total time $t = 6$ hours.

Number of half-lives, $n = \dfrac{t}{t_{1/2}} = \dfrac{6}{3} = 2$.

The amount of sucrose remaining is $N_t = \dfrac{N_0}{2^2} = \dfrac{N_0}{4}$.

Percentage of sucrose remaining $= \dfrac{N_t}{N_0} \times 100 = \dfrac{1}{4} \times 100 = 25$.

Answer: $25$