Let the initial density be $\rho$ and volume be $V$. With a rise in temperature $\Delta T$, the new volume $V' = V(1 + \gamma \Delta T)$. The new density $\rho' = \frac{m}{V'} = \frac{m}{V(1 + \gamma \Delta T)} = \rho (1 + \gamma \Delta T)^{-1} \approx \rho (1 - \gamma \Delta T)$ (since $\gamma \Delta T \ll 1$). The change in density is $\Delta \rho = \rho - \rho' = \rho - \rho (1 - \gamma \Delta T) = \rho \gamma \Delta T$. The fractional change in density is $\frac{\Delta \rho}{\rho} = \gamma \Delta T$. Given $\gamma = 5 \times 10^{-4}\text{ K}^{-1}$ and $\Delta T = 40^{\circ}\text{C} = 40\text{ K}$. Fractional change $= 5 \times 10^{-4} \times 40 = 200 \times 10^{-4} = 0.020$.