Answer
a) less than zero
b) greater than zero
c) $V_D=1.6V$
Work Step by Step
(a) We know that the current is driven in the clockwise direction in the circuit and there is a potential drop from point A to point B. Thus, if point A is grounded then the potential at point B is less than zero.
(b) We know that there is a potential drop from point C to point A. Thus, if point A is grounded then the potential at point C is greater than zero.
(c) We know that
$I=\frac{V_1-V_2}{R_{8.50}+R_{6.22}+R_{15.1}}$
$I=\frac{15.0V-11.5V}{8.50\Omega+6.22\Omega+15.1\Omega}=0.1174A$
Now the potential at point D is given as
$V_{D}=V_A+V_{15.0}-IR_{15.1}$
We plug in the known values to obtain:
$V_{D}=0+15.0V-(0.1174A)(15.1\Omega)$
$V_{D}=1.6V$
