At $10^\circ C$ the value of the density of a fixed mass of an ideal gas divided by its pressure is $x$. At $110^\circ C$ this ratio is -

Ideal gas equation: $PV = nRT = \dfrac{m}{M_0}RT$, where $M_0$ is the molar mass.
$P = \dfrac{m}{V}\cdot\dfrac{RT}{M_0} = \rho\,\dfrac{RT}{M_0}$
$\Rightarrow \dfrac{\rho}{P} = \dfrac{M_0}{RT}$, i.e. $\dfrac{\rho}{P} \propto \dfrac{1}{T}$
$T_1 = 10 + 273 = 283$ K and $T_2 = 110 + 273 = 383$ K
$\dfrac{(\rho/P)_2}{(\rho/P)_1} = \dfrac{T_1}{T_2} = \dfrac{283}{383}$
$\left(\dfrac{\rho}{P}\right)_2 = \dfrac{283}{383}\,x$