From the ideal gas equation, $pV=\mu RT$

$\Rightarrow p=\frac{\mu RT}{V}$

Given, $p{T}^2=$ constant

$\frac{\mu RT}{V}.T^2=$ constant

$T^3=kV$ ... (i)

Where, $k=$ constant.

On differentiating both sides, we get

$3T^{2}dT=kdV$...(ii)

From Equations (i) and (ii), we get

$\frac{3dT}{T}=\frac{dV}{V}\Rightarrow \frac{dV}{VdT}=\frac{3}{T}$

$\therefore$ Coefficient of volume expansion, $\frac{dV}{VdT}=\frac{3}{T}.$