According to question,


In a right angle triangle, in $\triangle A B C$
$$
\begin{aligned}
A C^2 & =A B^2+B C^2 \\
\Rightarrow \quad A C & =\sqrt{A B^2+B C^2}=\sqrt{\left(4 \times 10^{-2}\right)^2+\left(3 \times 10^{-2}\right)^2} \\
\Rightarrow \quad A C & =5 \times 10^{-2} \mathrm{~m}
\end{aligned}
$$

Initial electric potential energy of three charges,
$$
\begin{aligned}
U & =\frac{k q_1 q_2}{A B}+\frac{k q_1 q_3}{A C}+\frac{k q_2 q_3}{B C} \\
& =\frac{k\left(4 \times 3 \times 10^{-12}\right)}{4 \times 10^{-2}}+\frac{k\left(4 \times 5 \times 10^{-12}\right)}{5 \times 10^{-2}}+\frac{k\left(3 \times 5 \times 10^{-12}\right)}{3 \times 10^{-2}} \\
& =k\left[3 \times 10^{-10}+4 \times 10^{-10}+5 \times 10^{-10}\right] \\
& =9 \times 10^9 \times 12 \times 10^{-10} \\
& =108 \times 10^{-1}=10.8 \mathrm{~J}
\end{aligned}
$$

When three charges located of an equilateral triangle of side $3 \mathrm{~cm}$, the final potential energy of three charges system,
$$
\begin{gathered}
=\frac{k q_1 q_2}{A B}+\frac{k q_2 q_3}{B C}+\frac{k q_1 q_3}{A C} \\
=\frac{k\left(4 \times 3 \times 10^{-12}\right)}{3 \times 10^{-2}}+\frac{k\left(3 \times 5 \times 10^{-12}\right)}{3 \times 10^{-2}}+\frac{k\left(4 \times 5 \times 10^{-12}\right)}{3 \times 10^{-2}} \\
=\frac{9 \times 10^9}{3}\left[12 \times 10^{-10}+15 \times 10^{-10}+20 \times 10^{-10}\right]=14.1 \mathrm{~J}
\end{gathered}
$$

Hence, the work done in moving charges at points $A$ and $C, W=(14.1-10.8)=3.3 \mathrm{~J}$.