Answer
a) $V_0/R$
b) $10^5\;\rm V/m$
Work Step by Step
$$\color{blue}{\bf [a]}$$
The electric field strength $E_0$ at the surface of a charged sphere means at $r\geq R$.
Let's assume that the sphere is uniformly charged, so the electric field outside it is given by
$$E=\dfrac{1}{4\pi \epsilon_0}\dfrac{Q}{r^2}$$
So on its own surface, $r=R$
$$E_0=\dfrac{1}{4\pi \epsilon_0}\dfrac{Q}{R^2}$$
where the electric potential of a charged sphere is given by $V_0=\dfrac{1}{4\pi \epsilon_0}\dfrac{Q}{R}$.
Thus,
$$\boxed{E_0=\dfrac{V_0}{R}}$$
$$\color{blue}{\bf [b]}$$
We just need to plug the known into the boxed formula above.
$$ E_0=\dfrac{500}{0.5\times 10^{-2}} =\color{red}{\bf 1.0\times 10^5}\;\rm V/m$$
