Answer
$2.8\times 10^{7}\;\rm m/s$
Work Step by Step
We have an attractive electric force exerted on an electron by the positively charged glass sphere. This is similar to the gravitational force between the Earth and the Moon.
According to Newton's second law,
$$\sum F_r=F_E=ma_r$$
where $F_E=kQq/r^2$ and $a_r=v^2/r$
$$F_E=ma_r$$
$$\dfrac{kq_eQ}{r^{ \color{red}{\bf\not} 2}}=\dfrac{m_ev^2}{ \color{red}{\bf\not} r}$$
where $r$ here is the orbit radius, not the radius of the sphere, $r=R+h$ where $R$ is the radius of the sphere and $h$ is the height of the electron from the sphere's surface.
Solving for $v$,
$$v=\sqrt{\dfrac{kq_eQ}{r m_e}}$$
$$v=\sqrt{\dfrac{kq_eQ}{ (R+h)m_e}}$$
Plug the known;
$$v=\sqrt{\dfrac{(9\times 10^9)(1.6\times 10^{-19})(1\times 10^{-9})}{ (1+1)\times 10^{-3}(9.11\times 10^{-31})}}$$
$$v=\color{red}{\bf 2.8\times 10^{7}}\;\rm m/s$$
