Let $\alpha \text{ and }\beta$ be the roots of equation $p{x}^2+qx+r = 0$, $p\ne 0$. If $p, q, r$are in A.P. and $\frac{1}{\alpha }+\frac{1}{\beta }=4$, then the value of $\left|\alpha -\beta \right|$ is 

$\frac{1}{\alpha }+\frac{1}{\beta }=4$

   $\frac{\alpha +\beta }{\alpha \beta }=4$

$\therefore \frac{-\frac{q}{p}}{\frac{r}{p}}=4$

$\therefore -\frac{q}{r}=4$

$p, q, r$ are in A.P.

Let $p=a-d ; q=a, r=a+d$.

$\therefore \frac{-\left(a\right)}{a+d}=4 \Rightarrow -a=4a+4d \Rightarrow d=\frac{-5a}{4}$

$p=a+\frac{5a}{4}=\frac{9a}{4}; q=a, r=a-\frac{5a}{4}=-\frac{a}{4}$

$\therefore \frac{9}{4}{x}^2+\frac{4}{4}x-\frac{1}{4}=0$

$\therefore 9x^2+4x-1=0$

${\left|\alpha -\beta \right|}^2={\left(\alpha +\beta \right)}^2-4\alpha \beta$

           $={\left(\frac{-4}{9}\right)}^2-4\left(-\frac{1}{\text{9}}\right)$

           $=\frac{16}{81}+\frac{4}{9}=\frac{52}{81}$

$\therefore \left|\alpha -\beta \right|=\frac{2\sqrt{13}}{9}$.

Hence, the correct answer is $\frac{2\sqrt{13}}{9}$.