For a first order reaction,

$k=\frac{2.303}{t} \log \frac{[\mathrm{R}]_{0}}{[\mathrm{R}]}$

It is given that, t1/2 = 3.00 hours

Therefore, $k=\frac{0.693}{t_{1 / 2}}$

$=\frac{0.693}{3} \mathrm{~h}^{-1}$

= 0.231 h−1

Then, $0.231 \mathrm{~h}^{-1}=\frac{2.303}{8 \mathrm{~h}} \log \frac{[\mathrm{R}]_{0}}{[\mathrm{R}]}$

$\Rightarrow \log \frac{[\mathrm{R}]_{0}}{[\mathrm{R}]}=\frac{0.231 \mathrm{~h}^{-1} \times 8 \mathrm{~h}}{2.303}$

$\Rightarrow \frac{[\mathrm{R}]_{0}}{[\mathrm{R}]}=\operatorname{antilog}(0.8024)$

$\Rightarrow \frac{[R]_{0}}{[R]}=6.3445$

$\Rightarrow \frac{[\mathrm{R}]}{[\mathrm{R}]_{0}}=0.1576$ (approx)

$=0.158$

Hence, the fraction of sample of sucrose that remains after 8 hours is 0.158.