Resistance of the wire in terms of the diameter can be written as, $R=\frac{\rho L}{\pi {\displaystyle \frac{d^2}{4}}}$

Hence, percentage error relation will be, $\frac{∆R}{R}\times 100=\frac{∆L}{L}\times 100+\frac{2∆d}{d}\times 100$

Given: $\frac{∆L}{L}=0.1\%$ and $\frac{∆d}{d}=0.1\%$

Therefore, $\frac{∆R}{R}=0.3\%$