Answer
$4.2\times 10^{-10}\;\rm C$
Work Step by Step
We are given the potential difference of two points that are outside our charged bead, as we see in the figure below.
Hence, at $r\geq R$
$$V=\dfrac{kQ}{r}$$
Thus,
$$V_2-V_1=\dfrac{kQ}{r_2}-\dfrac{kQ}{r_1}$$
$$V_2-V_1=kQ\left(\dfrac{1}{r_2}-\dfrac{1}{r_1}\right)$$
Solving for $Q$;
$$Q=\dfrac{V_2-V_1}{k\left(\dfrac{1}{r_2}-\dfrac{1}{r_1}\right)} $$
Plug the known;
$$Q=\dfrac{-500}{(9\times 10^9)\left(\dfrac{1}{(5\times 10^{-3})}-\dfrac{1}{(3\times 10^{-3})}\right)} $$
$$Q=\color{red}{\bf 4.17\times 10^{-10}}\;\rm C$$
