Answer
$63\%$
Work Step by Step
By considering the mass of the solution, we can write
$M_{sol}=M_{water}+M_{eth}-(1)$
We know that,
$mass=density\times volume$
So (1)=>
$\rho_{s}V_{s}=\rho_{w}V_{w}+\rho_{e}V_{e}-(2)$
We can write,
$V_{s}=V_{w}+V_{e}=>V_{w}=V_{s}-V_{e}-(3)$
(3)=>(2),
$\rho_{s}V_{s}=\rho_{w}(V_{s}-V_{e})+\rho_{e}V_{e}$
$\frac{V_{e}}{V_{s}}=\frac{\rho_{s}-\rho_{w}}{\rho_{e}-\rho_{w}}$ ; Let's plug known values into this equation.
$\frac{V_{e}}{V_{s}}=\frac{1.073\times1000\space kg/m^{3}-1000\space kg/m^{3}}{1116\space kg/m^{3}-1000\space kg/m^{3}}\approx0.63$
The volume percentage of Ethylene glycol = $63\%$
