Answer
$V_B-V_A=-1.92MV$
Work Step by Step
The electric potential at point $A$ is given as
$VA=K\frac{q}{R}$
We plugin the known values to obtain:
$V_A=9\times 10^9\times \frac{16\times 10^{-6}}{0.03}$
$V_A=4800000V$
The electric potential at point $B$ is given as
$V_B=K\frac{q}{r}$
We plugin the known values to obtain:
$V_B=9\times 10^9\times \frac{16\times 10^{-6}}{\sqrt{(0.03)^2+(0.04)^2}}=2880000V$
Thus the potential difference is
$V_B-V_A=2880000-4800000$
$V_B-V_A=-1920000$
$V_B-V_A=-1.92MV$
