As one travels along the x -axis from x = -1 cm to x = +1 cm , the...

The $x$-component of the electric field is given by $E_{x} = -\dfrac{\partial V}{\partial x}$.

Since the potential decreases uniformly along the $x$-axis, we have $E_{x} = -\dfrac{\Delta V}{\Delta x} = -\dfrac{80 - 120}{1 - (-1)} = 20 \text{ V cm}^{-1}$.

The total electric field is $\vec{E} = E_{x} \hat{i} + E_{y} \hat{j} + E_{z} \hat{k}$.

The magnitude of the electric field is $E = \sqrt{E_{x}^2 + E_{y}^2 + E_{z}^2} = \sqrt{20^2 + E_{y}^2 + E_{z}^2}$.

Since the variation of potential along the $y$ and $z$ axes is not given, $E_{y}$ and $E_{z}$ are unknown and can be zero or non-zero.

Therefore, $E \ge 20 \text{ V cm}^{-1}$.

Thus, the magnitude of the electric field may be equal to $20 \text{ V cm}^{-1}$ (if $E_{y} = E_{z} = 0$) or it may be greater than $20 \text{ V cm}^{-1}$ (if $E_{y} \ne 0$ or $E_{z} \ne 0$).