Answer
$$V=-{2k\lambda}\ln[{\frac{s}{a}}]$$
$$ \mathbf{\vec \nabla} \mathbf{V}=-\frac{2k\lambda}{s}\hat s=-\vec E$$
Work Step by Step
We know that
Electric field due to long straight wire is
$$\vec E=\frac{2k\lambda}{s}\hat s$$
Potential(V) in the form of electric field
$$V=\int\limits_{a}^{s}-E. ds$$
In this case, we cannot set the reference point at infinity, since the charge itself extends to infinity.
$$V=\int\limits_{a}^{s}-[\frac{2k\lambda}{s}\hat s]. \vec ds$$
$$V=-\int\limits_{a}^{s}\frac{2k\lambda}{s} ds$$
$$V=-{2k\lambda}ln{s}\Big|_a^s \ $$
$$V=-{2k\lambda}\ln[{\frac{s}{a}}]$$
In this form, it is clear why a = 1 would be no good—likewise the other “natural” point, a = 0
$$ \mathbf{\vec \nabla} \mathbf{V}=\frac{\partial V}{\partial s} \hat s+ \frac{\partial V}{\partial \theta}\hat \theta + \frac{\partial V}{\partial \phi} \hat \phi $$
Potential is dependent only on s
So
$$ \mathbf{\vec \nabla} \mathbf{V}=\frac{\partial V}{\partial s}\hat s$$
$$ \mathbf{\vec \nabla} \mathbf{V}=-\frac{\partial ({2k\lambda}\ln[{\frac{s}{a}}]) }{\partial s}\hat s$$
$$ \mathbf{\vec \nabla} \mathbf{V}=-\frac{\partial ({2k\lambda}\ln[{\frac{s}{a}}]) }{\partial s}\hat s$$
$$ \mathbf{\vec \nabla} \mathbf{V}=-2k\lambda\frac{a}{s}\frac{1}{a}$$
The gradient of potential is
$$ \mathbf{\vec \nabla} \mathbf{V}=-\frac{2k\lambda}{s} $$
$$ \mathbf{\vec \nabla} \mathbf{V}=-\frac{2k\lambda}{s}\hat s=-\vec E$$