$The electric potential due to a point charge q at a distance [math]r[/math] is [math]V=\frac{1}{4\pi\epsilon_0}\left(\frac{q}{r}\right).[/math]$

$A point charge of 10^{-8} C is placed at the origin.$

$The distance of point A from the origin is [math]r_A=\sqrt{4^2+4^2+2^2}=6.[/math]$

$\Rightarrow \qquad The electric potential at [math]A[/math] is [math]V_A=\frac{1}{4\pi\epsilon_0}\left(\frac{q}{r_A}\right).[/math]$

$The distance of point B from the origin is [math]r_B=\sqrt{1^2+1^2+2^2}=3.[/math]$

$\Rightarrow \qquad The electric potential at [math]B[/math] is [math]V_B=\frac{1}{4\pi\epsilon_0}\left(\frac{q}{r_B}\right).[/math]$

$\Rightarrow \qquad The potential difference between points [math]A[/math] and [math]B[/math] is$

$\qquad V_B-V_A = \frac{1}{4\pi\epsilon_0}\left(\frac{q}{r_A}\right)-\frac{1}{4\pi\epsilon_0}\left(\frac{q}{r_B}\right)$

$\qquad = \frac{1}{4\pi\epsilon_0}\cdot q\left(\frac{1}{r_B}-\frac{1}{r_A}\right) = 9\times 10^9 \times 10^{-8}\times \left(\frac{1}{3}-\frac{1}{9}\right) = 20\,\,V.$