#### Answer

a.) The current in each wire is the same
b.) $J_1 > J_2$
c.) $\vec{E}_1 > \vec{E}_2$
d.) $(v_d)_1 > (v_d)_2$

#### Work Step by Step

a.) The current is the same in both wires because the same amount of charge flows through them. Recall that Kirchoff's junction law tells us that the amount of current flowing into a wire is the same amount of current flowing out. In this case, the same amount of current flowing into wire 1 must be the same amount of current flowing into wire 2 since they are connected.
b.) Recall that $\displaystyle J = \frac{I}{A_{\perp}}$. Since the current is the same in both wires, we can say that $J \displaystyle \propto \frac{1}{A_{\perp}}$. The radius of wire 2 is larger than wire 1, so wire 2 has a greater cross-sectional area, $A_{\perp}$. Because of this, wire 1 has a larger current density than wire 2.
$J_1 > J_2$
c.) Recall that $J = \sigma\vec{E}$, which means that $J$ is directly related to $\vec{E}$ (that is , $J \propto \vec{E}$). Since $J_1 > J_2$, $\quad \vec{E}_1 > \vec{E}_2$
d.) Finally, recall that $J = n_eev_d$ where $n_e$ is the electron density, $e$ is the charge of an electron, and $v_d$ is the drift speed of an electron. Since the wires are made of the same material, $n_e$ for each wire is the same. Also, each one has electrons flowing through it leaving us with, yet again, current density being directly proportional to drift speed ($J \propto v_d$). Therefore, since $J_1 > J_2$
$(v_d)_1 > (v_d)_2$