Answer
$V_R = -6V$
$P_I=12W$
Work Step by Step
The current goes from the negative to the positive for the resistor. Thus, we find:
$V=-IR = - (3A)(2\Omega) =\fbox{ -6V}$
We first need to find the voltage of the current source using Kirchhoff's Voltage Law:
$-10+6+v=0 \\ v-4=0 \\ v=4V$
Using this, we find power:
$P = VI = (4V)(3A) = \fbox{12 W}$
