The resistance of the wire is given as, $R=\frac{\rho l}{A}$

Therefore, $\rho =R\frac{A}{\mathcal{l}}$

$\rho =R\frac{\left(\pi r^2\right)}{\mathcal{l}}$

For percentage error, we can write $\frac{\Delta \rho }{\rho }=\frac{\Delta R}{R}+2\frac{\Delta r}{r}+\frac{\Delta \mathcal{l}}{\mathcal{l}}$

$\frac{\Delta \rho }{\rho }\times 100=\frac{\Delta R}{R}\times 100+2\frac{\Delta r}{r}\times 100+\frac{\Delta \mathcal{l}}{\mathcal{l}}\times 100$

$=\left(\frac{10}{100}+2\times \frac{0.05}{0.35}+\frac{0.2}{15}\right)\times 100$

$=\left(\frac{1}{10}+\frac{2}{7}+\frac{1}{75}\right)\times 100$

Hence, $\frac{\Delta \rho }{\rho }=39.9\%$