Potential at $x = 2$ due to the two charges placed at $x = 0$ and $x = 1$

$V= \large\frac{1}{4 \pi \in_0} \bigg[-\large\frac{1 \times 10^{-6}}{2} + \frac{1 \times 10^{-6}}{1} \bigg]V$

$\quad=4.5 \times 10^3 V$

Hence work done in bringing the charge from infinity will be $W = q_3( \Delta V) $

$\quad= q_3 (V-V_{\infty} = 1 \times 10^{-6} (4.5 \times 10^{3} -0)$

$\quad= 4.5 \times 10^{-3}\;J$

Hence A is the correct answer.