If the percentage errors in measuring the length and the diameter...

Resistance of the wire in terms of the diameter can be written as, $R = \dfrac{ρL}{\pi\dfrac{{d}^{2}}{4}}$

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

Given: $\dfrac{∆ L}{L} = 0 . 1$ and $\dfrac{∆ d}{d} = 0 . 1$

Therefore, $\dfrac{∆ R}{R} = 0 . 3$