Answer
a) $8.33\;\rm \mu C$
b) $3.32\;\rm MV/m$
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the electric potential of a charged sphere on its surface is given by
$$V =\dfrac{1}{4\pi \epsilon_0}\dfrac{Q}{R}$$
So
$$Q=(4\pi \epsilon_0)RV$$
Plugging the known
$$Q=\dfrac{(15\times 10^{-2})(500,000)}{9\times 10^9}$$
$$Q=\color{red}{\bf 8.33}\;\rm \mu C$$
$$\color{blue}{\bf [b]}$$
We know that the electric field strength outside a charged sphere is given by
$$E=\dfrac{Q}{A\epsilon_0}=\dfrac{Q}{4\pi r^2\epsilon_0}$$
where just outside the surface of the sphere means when $r=R$,
$$E=\dfrac{Q}{4\pi R^2\epsilon_0}$$
$$E=\dfrac{8.33\times 10^{-6}}{4\pi (0.15)^2((8.85\times 10^{-12})}$$
$$E=\color{red}{\bf 3.32\times 10^6}\;\rm V/m$$
