Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
According to the charge conservation principle, and since the current is given by $I=Q/t$, we can call it the conservation of the current principle, the current remains constant in the same wire.
Thus,
$$\boxed{I_1=I_2}$$
$$\color{blue}{\bf [b]}$$
The current density is given by
$$J=\dfrac{I}{A}$$
where $A$ is the cross-sectional area of the wire.
Hence,
$$\dfrac{J_1}{J_2}=\dfrac{I_1/A_1}{I_2/A_2}$$
where $I_1=I_2$,
$$\dfrac{J_1}{J_2}=\dfrac{A_2}{A_1}$$
and since $A_2\gt A_1$,
$$\boxed{J_1\gt J_2}$$
$$\color{blue}{\bf [c]}$$
We know that the electric field is given by
$$E=\dfrac{J}{\sigma}$$
where $\sigma$ is the conductivity
Hence,
$$\dfrac{E_1}{E_2}=\dfrac{J_1/\sigma_1}{J_2/\sigma_2}$$
where $\sigma_1=\sigma_2$ since the two parts are made of the same metal.
$$\dfrac{E_1}{E_2}=\dfrac{J_1 }{J_2 }$$
and since $J_1\gt J_2$,
$$\boxed{E_1\gt E_2}$$
$$\color{blue}{\bf [d]}$$
We know that current density is also given by
$$J =nev_{\rm d}$$
where $n$ is the electron density and it is constant for the same metal.
Hence,
$$v_{\rm d}=\dfrac{J}{ne}$$
$$\dfrac{(v_{\rm d})_1}{(v_{\rm d})_2}=\dfrac{J_1/ne}{J_2/ne}$$
$$\dfrac{(v_{\rm d})_1}{(v_{\rm d})_2}=\dfrac{J_1 }{J_2 }$$
and since $J_1\gt J_2$,
$$\boxed{{(v_{\rm d})_1}\gt {(v_{\rm d})_2}}$$
