R = (65 ± 1) $\Omega$, l = (5±0.1) mm and d = (10 ± 0.5) mm. Find error in calculation of resistivity.   

Given, R = 65 $\Omega$, $∆$R = I $\Omega$,
            l = 5 x 10^-3 m, $∆$l = 0.1 X 10^-3 m,
           d = 10 x 10^-3 m and $∆$d = 0.5 x 10^-3 m

$\because$   Resistivity, $\rho =\frac{RA}{l}\text{ or }\rho =\frac{R\pi (d/2{)}^2}{l}=\frac{\pi R{d}^2}{4l}$

$\therefore$ $\frac{\mathrm{\Delta }\rho }{\rho }=\frac{\mathrm{\Delta }R}{R}+2\frac{\mathrm{\Delta }d}{d}+\frac{\mathrm{\Delta }l}{l}$

$\Rightarrow \frac{\mathrm{\Delta }\rho }{\rho }=\frac{1}{65}+2\left(\frac{0.5\times {10}^{-3}}{10\times {10}^{-3}}\right)+\frac{0.1\times {10}^{-3}}{5\times {10}^{-3}}$

$\Rightarrow \frac{\mathrm{\Delta }\rho }{\rho }=0.0153+0.1+0.02\Rightarrow \frac{\mathrm{\Delta }\rho }{\rho }\approx 0.1353$

So, error in calculation of resistivity is 13.5% = 13%.