Answer
$4.9\;\rm cm/s, \;9.8\;cm/s$
Work Step by Step
Since the two beads have unlike charges, they will toward each other when they are released.
Let's assume that the two beads are rolling without friction. This means that the system is isolated which means that the energy and momentum are conserved.
$$E_i=E_f$$
$$K_{iC}+K_{iD}+U_{i}=K_{fC}+K_{fD}+U_{f}$$
The two beads are released from rest, so their initial kinetic energy is zero.
$$0+0+U_{i}=K_{fC}+K_{fD}+U_{f}$$
$$ \dfrac{k_eq_Cq_D}{r_i}=\frac{1}{2}m_Cv_C^2+\frac{1}{2}m_Dv_D^2+\dfrac{k_eq_Cq_D}{r_f}$$
The distance between the centers of the two beads when they collide is equal to the diameter of one of them, so $r_f=D$
$$ \dfrac{k_eq_Cq_D}{r_i}=\frac{1}{2}m_Cv_C^2+\frac{1}{2}m_Dv_D^2+\dfrac{k_eq_Cq_D}{D}$$
$$ m_Cv_C^2+ m_Dv_D^2=\dfrac{2k_eq_Cq_D}{r_i} -\dfrac{2k_eq_Cq_D}{D}$$
$$m_Cv_C^2+ m_Dv_D^2=2k_eq_Cq_D\left[\dfrac{1}{r_i} -\dfrac{1}{D}\right]\tag 1$$
Now Let's deal with momentum conservation,
$$p_i=p_f$$
$$m_Cv_{iC}+ m_Dv_{iD}=m_Cv_{C}+ m_Dv_{D}$$
$$0+ 0=m_C v_{C}+ m_Dv_{D}$$
$$v_C=\dfrac{- m_Dv_{D}}{m_C }\tag 2$$
Plug into (1),
$$m_C\left[\dfrac{- m_Dv_{D}}{m_C }\right]^2+ m_Dv_D^2=2k_eq_Cq_D\left[\dfrac{1}{r_i} -\dfrac{1}{D}\right] $$
$$ \dfrac{ m_D^2v_{D}^2}{m_C } + m_Dv_D^2=2k_eq_Cq_D\left[\dfrac{1}{r_i} -\dfrac{1}{D}\right] $$
$$v_{D}^2\left[ \dfrac{ m_D^2}{m_C } + m_D \right]=2k_eq_Cq_D\left[\dfrac{1}{r_i} -\dfrac{1}{D}\right] $$
$$v_{D}^2\left[ \dfrac{ m_D^2 + m_C m_D}{m_C } \right]=2k_eq_Cq_D\left[\dfrac{1}{r_i} -\dfrac{1}{D}\right] $$
$$v_{D} =\sqrt{\dfrac{2k_eq_Cq_D\left[\dfrac{1}{r_i} -\dfrac{1}{D}\right] }{m_D\left[ \dfrac{ m_D + m_C }{m_C } \right]}}$$
Plug the known;
$$v_{D} =\sqrt{\dfrac{2(9\times 10^9)(2\times 10^{-9})(-1 \times 10^{-9}) \left[\dfrac{1}{10\times 10^{-3}} -\dfrac{1}{2\times 10^{-3}}\right] }{(2\times 10^{-3})\left[ \dfrac{ 2 + 1}{1} \right]}}$$
$$v_D=\bf 0.04899\;\rm m/s\approx\color{red}{\bf{4.9}} \;\rm cm/s $$
Plug into (2),
$$v_C=\dfrac{- (2)(4.9)}{1} $$
$$v_C\approx \color{red}{\bf- 9.8} \;\rm m/s $$