Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
Since the surface is frictionless, the mechanical energy is conserved.
$$E=K+U_e$$
The system is initially at rest, so $K=0$
$$E=U_e=\dfrac{kq_1q_2}{r}$$
where $q_1=q_2$
$$E =\dfrac{kq^2}{r}$$
Plug the known;
$$E =\dfrac{(9\times 10^9)(2\times 10^{-6})^2}{(5\times 10^{-2})}$$
$$E=\color{red}{\bf 0.72}\;\rm J\tag 1$$
$$\color{blue}{\bf [b]}$$
Let's draw the force diagram exerted on on cube of both, as shown below.
The 2-g cube is a rest, the whole system actually is at rest, so the net force exerted on it in the $x$-direction is zero.
So,
$$\sum F_x=F_e-T=m_1a_x=m_1(0)=0$$
Hence,
$$T=F_e=\dfrac{kq^2}{r^2}$$
Plug the known;
$$T= \dfrac{(9\times 10^9)(2\times 10^{-6})^2}{(5\times 10^{-2})^2}$$
$$T=\color{red}{\bf 14.4}\;\rm N$$
$$\color{blue}{\bf [c]}$$
When the string is cut $T=0$, and when the two cubes are far apart, then $r\rightarrow \infty$.
Noting that the energy is still conserved, and so the momentum,
$$E=K_f+U_{ef}$$
where $U_{ef}=0$ since $r\rightarrow \infty$,
$$E=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2$$
Plug from (1),
$$0.72=\frac{1}{2}\left[ m_1v_1^2+ m_2v_2^2\right]$$
$$1.44= m_1v_1^2+ m_2v_2^2 $$
Recalling that $m_2=2m_1$
$$1.44= m_1v_1^2+ 2m_1v_2^2 \tag 2 $$
The momentum is conserved,
$$p_f=p_i$$
$$p_f=0$$
$$m_1v_1+m_2v_2=0$$
$$m_1v_1+2m_1v_2=0$$
Hence,
$$ v_1=-2 v_2 \tag 3$$
Plug into (2),
$$1.44= m_1(-2 v_2)^2+ 2m_1v_2^2 $$
$$1.44= 4m_1 v_2^2+ 2m_1v_2^2 $$
$$1.44= 6m_1v_2^2 $$
Thus,
$$v_2=\pm\sqrt{\dfrac{1.44}{6(2\times 10^{-3})}}$$
$$v_2=\color{red}{\bf -10.95}\;\rm m/s$$
The negative sign is due to the direction of velocity which is toward the left.
Plugging into (3),
$$v_2=\color{red}{\bf +21.9}\;\rm m/s$$
The positive sign is due to the direction of velocity which is toward the right.
