Answer
See the detailed answer below.
Work Step by Step
According to Gauss’s law,
$$\oint \vec E d \vec A=\dfrac{Q_{in}}{\epsilon_0}$$
Hence,
$$EA_{sphere}=\dfrac{Q_{in}}{\epsilon_0}$$
So the electric field is given by
$$E=\dfrac{Q_{in}}{A_{sphere}\epsilon_0}$$
$$\boxed{E=\dfrac{Q_{in}}{4\pi r^2\epsilon_0}}$$
where $r$ is the radius of the Gaussian sphere.
We will use this formula to find the electric field for both cases.
In the figure below we have 4 dashed circles which represent the 4 Gaussian spheres.
$$\color{blue}{\bf [a]}$$
$\bullet$ when $r\lt a$, see circle 1 in the figure below.
$$E=\dfrac{Q_{in}}{4\pi r^2\epsilon_0}$$
where $Q_{in}$ is the charge enclosed inside the Gaussian surface inside the ball. The charge inside the ball is uniformly distributed, so
$$\rho_{e}=\dfrac{-Q}{V}=\dfrac{-Q}{\frac{4}{3}\pi a^3}$$
Hence,
$$Q_{in}=\rho_e \frac{4}{3}\pi r^3=\dfrac{-Q}{\frac{4}{3}\pi a^3}\left( \frac{4}{3}\pi r^3\right)=\dfrac{-Qr^3}{a^3} $$
Thus,
$$E=\dfrac{-Qr^3}{4\pi r^2a^3\epsilon_0}$$
$$E=-\dfrac{ 1 }{4\pi \epsilon_0}\dfrac{Qr}{ a^3}$$
$$\boxed{\vec E=-\dfrac{ 1 }{4\pi \epsilon_0}\dfrac{Qr}{ a^3}}\tag{Inward}$$
$$\color{blue}{\bf [b]}$$
$\bullet$ when $a\lt r\lt b$, see circle 2 in the figure below.
$$E=\dfrac{Q_{in}}{4\pi r^2\epsilon_0}$$
where $Q_{in}=-Q$
$$E=\dfrac{-Q}{4\pi r^2\epsilon_0}$$
Hence,
$$\boxed{\vec E=-\dfrac{ 1}{4\pi \epsilon_0}\dfrac{Q}{r^2}\;\hat r}\tag{Inward}$$
$$\color{blue}{\bf [c]}$$
$\bullet$ when $b\lt r\lt c$, see circle 3 in the figure below.
$$E=\dfrac{Q_{in}}{4\pi r^2\epsilon_0}$$
where $Q_{in}=-Q+Q=0$
where $+Q$ is the charge of the inner surface of the hollow metal sphere due to polarization due to the negative charge of the ball.
Also, we know that the electric field inside a conductor is zero.
$$E=\dfrac{-Q}{4\pi r^2\epsilon_0}$$
Hence,
$$\boxed{\vec E=\bf0\;\rm N/C}$$
$$\color{blue}{\bf [d]}$$
$\bullet$ when $r\gt c$, see circle 4 in the figure below.
$$E=\dfrac{Q_{in}}{4\pi r^2\epsilon_0}$$
where $Q_{in}=+2Q-Q=Q$
$$E=\dfrac{Q}{4\pi r^2\epsilon_0}$$
Hence,
$$\boxed{\vec E=\dfrac{ 1}{4\pi \epsilon_0}\dfrac{Q}{r^2}\;\hat r}\tag{Outward}$$
