Answer
(a) greater than
(b) III
Work Step by Step
(a) We know that $E=-\frac{\Delta V}{\Delta s}$. As E points towards the right, the voltage difference is $V_3-V_1<0$. This shows that potential 1 greater than potential 3 because the electric potential decreases in the direction of the electric field.
(b) Option III is correct. That is, the electric field is more intense at point 2 than at point 1, which means that the equipotential surfaces are more closely spaced in the region.