$\frac{1 \Delta \mathrm{R}}{\mathrm{R}}=\frac{1 \Delta \ell}{\ell}+\frac{2 \Delta \mathrm{r}}{\mathrm{r}}+\frac{\Delta \rho}{\rho}$
Maximum error is least count of an instrument
$\therefore \quad 1.5 \%$






Reason : At high voltage supply power losses are less.