Answer
$W = -356 \times 10^{-3}\,\text{J}$
Work Step by Step
From the triangle below we could get the distance between the two charges when the $q_2$ at the final location by
\begin{gather*}
r_b^2 = 0.250^2 + 0.250^2 \\
r_b = \sqrt{0.250^2 + 0.250^2}\\
r_b = 0.354 \,\text{m}
\end{gather*}
Also, we are given $r_a $ = 0.150 m. The work done on $q_2$ is given by
\begin{align*}
W &= \dfrac{1}{4\pi \epsilon_o} (q_1q_2) \left[\dfrac{1}{r_a} - \dfrac{1}{r_b} \right]\\
&= (9\times 10^{9} \mathrm{~N\cdot m^2/C^2})(2.4\times 10^{-6}) (-4.3\times 10^{-6}) \left[\dfrac{1}{0.150 \,\text{m}} - \dfrac{1}{0.354 \,\text{m}} \right]\\
&=\boxed{ -356 \times 10^{-3} \,\text{J}}
\end{align*}
