Answer
See work below.
Work Step by Step
a. Consider a mass m of gasoline with volume $V_0$ at $0^{\circ}$. Its density is $\rho_0=\frac{m}{V_0}$.
At a hot temperature of $33^{\circ}$, the volume grows to $V_0(1+\beta \Delta t)$.
Calculate the density.
$$\rho=\frac{m}{V_0(1+\beta \Delta t)}=\frac{\rho_0}{1+\beta \Delta t}$$
$$\rho=\frac{0.68\times10^3 kg/m^3}{1+(950\times10^{-6}/C^{\circ})(33C^{\circ})}=0.6593\times10^3 kg/m^3$$
$$=0.66\times10^3 kg/m^3$$
b.
$$\frac{(0.6593-0.68)\times10^3 kg/m^3}{0.68\times10^3 kg/m^3}=-3.0\%$$
