Answer
$\frac{\Delta V}{V_0} = 3~\alpha~\Delta T$
$\beta = 3~\alpha$
Work Step by Step
Let the original area $V_0 = s_0^3$
We can find an expression for the new volume after the material expands:
$V = (s_0+\Delta s)^3$
$V = s_0^3+3s_0^2~\Delta s+3s_0 (\Delta s)^2+(\Delta s)^3$
$V \approx s_0^3+3s_0^2~\Delta s$
We can find an expression for the ratio $\frac{\Delta V}{V_0}$:
$\Delta V = V-V_0$
$\Delta V = (s_0^3+3s_0^2~\Delta s)-s_0^3$
$\Delta V = 3s_0^2~\Delta s$
$\Delta V = 3s_0^2~(\alpha~\Delta T~s_0)$
$\Delta V = 3s_0^3~\alpha~\Delta T$
$\Delta V = 3~V_0~\alpha~\Delta T$
$\frac{\Delta V}{V_0} = 3~\alpha~\Delta T$
Since $\frac{\Delta V}{V_0} = \beta~\Delta T$, we can see that $\beta = 3~\alpha$
