A metal ball of radius 4.0 cm sits at the centre of a hollow metal shell of inner radius 9.0 cm…

Step 1. The capacitance. $C = 4\pi\varepsilon_0\dfrac{ab}{b - a} = 4\pi \times 8.85 \times 10^{-12} \times \dfrac{0.040 \times 0.090}{0.050} = 8.01 \times 10^{-12}\ \text{F}$, that is $8.01$ pF.

Step 2. The charge. $Q = CV = 8.01 \times 10^{-12}\ \text{F} \times 150\ \text{V} = 1.20 \times 10^{-9}\ \text{C}$, that is $1.20$ nC.

Step 3. The field at the ball. Between the ball and the shell the field is that of the ball alone. At its surface $E = \dfrac{kQ}{a^2} = \dfrac{9.0 \times 10^{9} \times 1.20 \times 10^{-9}}{(0.040)^2} = 6.75 \times 10^{3}\ \text{V/m}$, that is $6.75$ kV/m.