where $\bar{\nabla}=\hat{i} \frac{\partial}{\partial x}+\hat{j} \frac{\partial}{\partial y}+\hat{k} \frac{\partial}{\partial z}$

$\therefore \quad \vec{E}=-\left[\hat{i} \frac{\partial V}{\partial x}+\hat{j} \frac{\partial V}{\partial y}+\hat{k} \frac{\partial V}{\partial z}\right]$

Here, $V=4 x^{2} \quad \therefore \quad \vec{E}=-8 x \hat{i}$

The electric field at point $(1,0,2)$ is

$\vec{E}_{(1,0,2)}=-8 \hat{i}\,Vm^{-1}$

So electric field is along the negative $X$ -axis.