Answer
a. $44C^{\circ}$
b. The bullet doesn't melt
Work Step by Step
a. There is more KE in the beginning; half of the KE lost goes toward heating the bullet.
$$\frac{1}{2}(\frac{1}{2}m(v_i^2-v_f^2))=mc_{lead}\Delta T$$
$$\Delta T=\frac{(v_i^2-v_f^2)}{4 c_{lead}}=\frac{(220m/s)^2-(160m/s)^2}{4 (130 J/(kg\cdot C^{\circ}))}=44C^{\circ}$$
b. The final temperature of the bullet is far below the melting temperature of lead. The bullet doesn’t melt at all.