Answer
$V_A-V_B=-4500V$
Work Step by Step
We know that $V_A=K\frac{q}{d_1}$.
First, we find $V_A$:
$V_A=9\times 10^9\times \frac{1\times 10^{-6}}{2}=4500V$
Next, we find $V_B$:
$V_B=K\frac{q}{d_2}$
$V_B=9\times 10^9\times \frac{1\times 10^{-6}}{1}=9000V$
Now,
$V_A-V_B=4500-9000$
$V_A-V_B=-4500V$