Work done by external agent : $$ \mathrm{W}_{\mathrm{ext}}=\Delta \mathrm{U} ; $$ $\Delta \mathrm{U} \rightarrow$ Change in potential energy in taking the charge from initial to final configuration $$ \Rightarrow \quad W_{e x t}=\frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r_f}-\frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r_i} $$Now, $r_f=\sqrt{(2-0)^2+(2-0)^2+(1-0)^2}=3 \mathrm{~m}$ $$ \begin{aligned} r_{\mathrm{i}} & =\sqrt{(4-0)^2+(4-0)^2+(2-0)^2}=6 \mathrm{~m} \\ \therefore \quad W_{\text {ext }} & =\left(9 \times 10^9\right) \times\left(10^{-8} \times 2 \times 10^{-6}\right)\left[\frac{1}{3}-\frac{1}{6}\right] \\ & =3 \times 10^{-5} \\ & =30 \times 10^{-6} \mathrm{~J} \end{aligned} $$ Correct Option (3)