Answer
$1.06\;\rm cm/s$, $1.77\;\rm cm/s$
Work Step by Step
Since the two beads are negatively charged, they will move away from 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.
$$E_i=E_f$$
$$K_{iA}+K_{iB}+U_{i}=K_{fA}+K_{fB}+U_{f}$$
The two beads are released from rest, so their initial kinetic energy is zero.
$$0+0+U_{i}=K_{fA}+K_{fB}+U_{f}$$
$$ \dfrac{k_eq_Aq_B}{r_i}=\frac{1}{2}m_Av_A^2+\frac{1}{2}m_Bv_B^2+\dfrac{k_eq_Aq_B}{r_f}$$
The two beads will reach the maximum speed when the distance between their centers is approaching infinity, $r_f=\infty$.
$$ \dfrac{k_eq_Aq_B}{r_i}=\frac{1}{2}m_Av_A^2+\frac{1}{2}m_Bv_B^2+0$$
$$ \dfrac{k_eq_Aq_B}{r_i}=\frac{1}{2}m_Av_A^2+\frac{1}{2}m_Bv_B^2\tag 1$$
Now Let's deal with momentum conservation,
$$p_i=p_f$$
$$m_Av_{iA}+ m_Bv_{iB}=m_Av_{A}+ m_Bv_{B}$$
$$0+ 0=m_A v_{A}+ m_Bv_{B}$$
$$v_A=\dfrac{- m_Bv_{B}}{m_A }\tag 2$$
Plug into (1),
$$ \dfrac{k_eq_Aq_B}{r_i}=\frac{1}{2}m_A\left[\dfrac{- m_Bv_{B}}{m_A }\right]^2+\frac{1}{2}m_Bv_B^2 $$
$$ \dfrac{k_eq_Aq_B}{r_i}=\frac{1}{2}m_A \dfrac{ m_B^2v_{B}^2}{m_A^2 }+\frac{1}{2}m_Bv_B^2 $$
$$ \dfrac{2k_eq_Aq_B}{r_i}= \dfrac{ m_B^2v_{B}^2}{m_A }+ m_Bv_B^2 $$
Hence,
$$ \dfrac{2k_eq_Aq_B}{r_i}=\left[ \dfrac{ m_B^2 }{m_A }+ m_B\right]v_B^2 $$
$$ \dfrac{2k_eq_Aq_B}{r_i}=\left[ \dfrac{ m_B^2+m_Am_B }{m_A }\right]v_B^2 $$
$$v_B=\sqrt{ \dfrac{2k_eq_Aq_B\;m_A}{r_im_B(m_B+m_A)}}$$
Plug the known;
$$v_B=\sqrt{ \dfrac{2(9\times 10^9)(-5\times 10^{-9})(-10\times 10^{-9}) (0.015)}{(0.12)(0.025)(0.025+0.015)}}$$
$$v_B=\bf{0.01061} \;\rm m/s\approx \color{red}{\bf 1.06}\;\rm cm/s$$
Plug into (2),
$$v_A=\dfrac{- (0.025)(0.01061)}{0.015} $$
$$v_AB=\bf{- 0.0177} \;\rm m/s\approx \color{red}{\bf -1.77}\;\rm cm/s$$
