Explanation

(c) On stretching, volume ($V$) remains constant. So
$V = Al\ or\ l = V/A.$
Now, $R = \frac{\rho l}{A} = \frac{\rho V}{A^{2}} = \frac{\rho V}{\pi^{2}D^{4}/16} = \frac{16\rho V}{\pi^{2}D^{4}}$
Taking logarithm of both the side and differentiating it we get
$\frac{\mathrm{\Delta}R}{R} = - 4\frac{\mathrm{\Delta}D}{D}or\frac{\mathrm{\Delta}R}{R} = - 4 \times 0.25 = 1.0\%$