Given that,

Relative error of height $\left(\frac{\delta h}{h}\right)=0.02$

Relative error of radius $\left(\frac{\delta r}{r}\right)=0.02$

Volume of the cone, $V=\frac{1}{3}\pi r^{2}h$

Applying log on both sides,

$\log V=\log \left(\frac{\mathrm{\pi }}{3}{r}^{2}h\right)$

$\Rightarrow \log V=\log \frac{\mathrm{\pi }}{3}+\log r^2+\log h$

$\Rightarrow \frac{1}{V}\delta V=0+2·\frac{\delta r}{r}+\frac{\delta h}{h}$

$\Rightarrow \frac{\delta V}{V}=2\left(0.02\right)+0.02=0.06$

Percentage error $=\frac{\delta V}{V}\times 100=0.06\times 100=6\%$

$\therefore$The percentage error in volume $=6$.