The work done $$(muJ)$$ in moving a point charge $$Q=-16muC$$ from  x=0 to x=3 in a field $\overline{E}=2x\mathit{i}$  V/m is most nearly:

A.   288
B.   144
C.   96
D.   48

Work is a measure of the total energy transfer that occurs when an object is acted upon by an external force over a specified distance. 

Since our electric field is non-uniform and will change depending on location, we can express work done in moving the point charge through the following integral equation:

$W ={\int }_A^{B}q\overrightarrow{E}·d\overrightarrow{s}$

The dot product selects only the radial component of our electric field.

Plugging our given values from the problem to set up our integral:

$$\begin{gathered} W ={\int }_0^3(-16·{10}^{-6})\left(2x\mathit{i}\right)·dx \\ W =(-16·{10}^{-6}){\int }_0^3\left(2x\right)dx \\ W =(-16·{10}^{-6})\left({\overline{)\frac{2x^2}{2}}}_0^3\right) \\ W =(-16·{10}^{-6})(\frac{2{\left(3\right)}^2}{2}-\frac{2{\left(0\right)}^2}{2}) \\ W =(-16·{10}^{-6})\left(9\right) \\ W =-144\mu J \end{gathered}$$

Work itself is a scalar quantity, and the negative difference in potential between starting and ending coordinates simply means that the work is being done against the electric field. 

Therefore, our final answer is  B, 144 uJ.