Answer
$0.49\;\rm m/s$
Work Step by Step
Let's assume that the 4 spheres are rolling in a frictionless surface which means that the energy and the momentum of the system are conserved.
$$E_i=E_f$$
$$U_i+K_i=U_f+K_f$$
where $K_i=0$ since the 4 spheres released from rest, and $U_f=0$ since $r\rightarrow \infty$.
$$U_i=K_f$$
$$[U_{12}+U_{23}+U_{34}+U_{41}]+[U_{13}+U_{24}]=K_1+K_2+K_3+K_4$$
The 4 spheres are identical in mass and in charge. This means that they must have the same speed when they are far apart.
$$4\left[ \dfrac{kq^2}{r_1}\right]+2\left[ \dfrac{kq^2}{r_2}\right]=4\left(\frac{1}{2}mv_f^2\right)$$
where $r_1=r_{12}=r_{23}=r_{34}=r_{31}=1$ cm, and $r_2=r_{13}=r_{24}=\sqrt{1^2+1^2}=\sqrt2$ cm.
$$ \dfrac{4kq^2}{0.01} +2\left[ \dfrac{100kq^2}{\sqrt2}\right]=2mv_f^2 $$
Hence,
$$v_f^2=\dfrac{kq^2}{m}\left[\dfrac{2 }{0.01} + \dfrac{100 }{\sqrt2}\right] $$
$$v_f=\sqrt{\dfrac{kq^2}{m}\left[\dfrac{2 }{0.01} + \dfrac{100 }{\sqrt2}\right] }$$
Plug the known;
$$v_f=\sqrt{\dfrac{(9\times 10^9)(10\times 10^{-9})^2}{(1\times 10^{-3})}\left[\dfrac{2 }{0.01} + \dfrac{100 }{\sqrt2}\right] }$$
$$v_f=\color{red}{\bf 0.49}\;\rm m/s$$
