Answer
$3.04\times 10^{-11}$
Work Step by Step
First of all, we need to find the number of electrons in such a sphere of copper. so we need to find the number of atoms in this sphere.
We know, from density law, that $\rho =m/V$
Hence, the mass of one sphere is
$$m=\rho V=\rho \left( \frac{4}{3}\pi r^3\right)$$
$$m= \frac{4}{3}\pi r^3 \rho$$
And we know that the number of atoms is given by
$$N=\dfrac{m}{M}$$
where $M$ is the atomic mass,
$$N=\dfrac{ \frac{4}{3}\pi r^3\rho}{M}$$
Plug the known;
$$N=\dfrac{ \frac{4}{3} \pi (1\times 10^{-3})^3(8920)}{(63.5\times 1.66\times 10^{-27})} $$
$$N_{Cu}=\bf 3.5446\times 10^{20}\;\rm atom $$
Now we know that the number of electrons or protons in one copper atom is 29.
Hence,
$$N_e=29 N_{Cu}=\bf 1.0279\times 10^{22}\;\rm electron\tag 1$$
This is the number of electrons, or the number of protons, in this sphere.
Now we need to find the number of electrons removed from the ball to charge it by a net charge of $Q=+50$ nC.
$$N_{e,\rm removed}=\dfrac{Q}{q_{e^-}}=\dfrac{(50\times 10^{-9})}{(1.6\times 10^{-19})}$$
$$N_{e,\rm removed}=\bf 3.125\times 10^{11}\tag 2$$
Hence, the fraction of electrons removed from the copper ball is given by dividing (2) by (1),
$$\dfrac{N_{e,\rm removed}}{N_e}=\dfrac{3.125\times 10^{11}}{1.0279\times 10^{22}}=\color{red}{\bf 3.04\times 10^{-11}}$$
