Answer
$$ \dfrac{\Delta l}{l_0}=\dfrac{\Delta R}{2R_0} $$
See the detailed answer below.
Work Step by Step
The author told us that the volume of the wire is assumed to remain constant
Thus,
$$V=A_0l_0=Al\tag 1$$
whereas $A$ is its cross-sectional area and $l$ is its length.
We know that the resistance of a wire s given by
$$R=\dfrac{\rho_el}{A}$$
We know from (1), that $A=\dfrac{V}{l}$. Plugging that into the previous formula.
$$R=\dfrac{\rho_el^2}{V}\tag 2$$
So, initially, the resistance is given by:
$$R_0=\dfrac{\rho_el_0^2}{V_0}\tag 3$$
We must subtract (3) from (2) to find $\Delta R$.
$$\Delta R=R-R_0=\dfrac{\rho_el^2}{V}-\dfrac{\rho_el_0^2}{V_0}$$
Noting that the volume remains constant, so $V_0=V$
Also, $l=l_0+\Delta l$ and we assume that $\Delta l \lt\lt l_0$
$$\Delta R= \dfrac{\rho_el^2-\rho_el_0^2}{V_0}=\dfrac{\rho_e(l_0+\Delta l)^2-\rho_el_0^2}{V_0}$$
$$\Delta R =\dfrac{\rho_e(l_0^2+2l_0\Delta l+\overbrace{ \Delta l^2}^{0} ) -\rho_el_0^2}{V_0}$$
Since $\Delta l\lt \lt l_0$, so $\Delta l^2\approx 0$
$$\Delta R =\dfrac{\rho_e(l_0^2+2l_0\Delta l ) -\rho_el_0^2}{V_0}=\dfrac{\rho_e l_0^2+2\rho_el_0\Delta l -\rho_el_0^2}{V_0}$$
$$\Delta R = \dfrac{2\rho_el_0\Delta l }{V_0}=2 \Delta l\left( \dfrac{ \rho_el_0 }{V_0}\right)\cdot \dfrac{l_0}{l_0}$$
$$\Delta R = \dfrac{ 2 \Delta l }{l_0}\;\;\overbrace{ \left( \dfrac{ \rho_el_0^2 }{V_0}\right) }^{R_0} $$
$$\Delta R = \dfrac{ 2 \Delta l }{l_0}R_0$$
Thus,
$$\boxed{\dfrac{\Delta l}{l_0}=\dfrac{\Delta R}{2R_0}}$$
