The acceleration of an electron due to the mutual attraction...

Recall the formula of electrostatic force interns of charge on electron and septation between them, $F = K\dfrac{{e}^{2}}{{r}^{2}}$, now use Newton's second law, $ma = K\dfrac{{e}^{2}}{{r}^{2}}$

$⇒ a = K\dfrac{{e}^{2}}{m{r}^{2}}$,

so, the acceleration, $a = 9 \times {10}^{9}\dfrac{{\left(1 . 6 \times {10}^{- 19}\right)}^{2}}{{\left(1 . 6 \times {10}^{- 10}\right)}^{2}\left(9 \times {10}^{- 31}\right)}$

$⇒ a = {10}^{- 29} \times {10}^{51} = {10}^{22}m{s}^{- 2}$