A small metal ball of radius 0.060 m carries +15 nC…

Step 1. The inner surface. The field is zero on a Gaussian sphere inside the metal, so it must enclose zero charge. It encloses the ball, $+15$ nC. So the inner surface carries $-15$ nC.

Step 2. The outer surface. The shell is neutral, so the two surfaces add to zero. The outer surface carries $+15$ nC.

Step 3. The field at 0.09 m. A Gaussian sphere here holds only the ball, $+15$ nC. $E = \dfrac{Q}{4\pi\varepsilon_0 r^2} = \dfrac{15 \times 10^{-9}\ \text{C}}{4\pi \times 8.85 \times 10^{-12}\ \text{C}^2/(\text{N m}^2) \times (0.09\ \text{m})^2} = 1.67 \times 10^4\ \text{N/C}$, that is $16.7$ kN/C.

Step 4. The field at 0.30 m. A Gaussian sphere here holds $+15 - 15 + 15 = +15$ nC. $E = \dfrac{15 \times 10^{-9}}{4\pi \times 8.85 \times 10^{-12} \times (0.30)^2} = 1.50 \times 10^3\ \text{N/C}$, that is $1.50$ kN/C.

Step 5. The density on the outer surface. $\sigma = \dfrac{Q}{4\pi R^2} = \dfrac{15 \times 10^{-9}\ \text{C}}{4\pi \times (0.15\ \text{m})^2} = 5.31 \times 10^{-8}\ \text{C/m}^2$, that is $53.1$ nC/m².