Answer
$60.3\%$
Work Step by Step
Let's apply Hooke's law when only the weight of the block compresses the spring.
$mg=kx-(1)$
Let's apply Hooke's law when the block is inside the water.
$F_{B}-mg=2kx-(2)$
(1)=>(2),
$F_{B}=3mg-(3)$
According to the principle of Archimedes', we can write,
$F_{B}=V_{B}\rho_{w}g-(4)$ ; $V_{B}$ is volume of the block.
(3)=(4)=>
$V_{B}\rho_{w}g=3mg=>V_{B}=\frac{3m}{\rho_{w}}=\frac{3(8\space kg)}{1000\space kg/m^{3}}=0.024\space m^{3}$
We can get the volume of wood in the block by,
$V_{W}=\frac{m}{\rho}=\frac{8\space kg}{840\space kg/m^{3}}=9.52\times10^{-3}m^{3}$
The volume of the hollow = $V_{B}-V=(0.024-0.00952)\space m^{3}=1.45\times10^{-2}m^{3}$
Percentage of the hollow = $\frac{1.45\times10^{-2}m^{3}}{0.024\space m^{3}}\times100\%\approx60.3\%$
