Consider that an electron is revolving in an excited state of...

The electrostatic force between the nucleus and the electron provides the necessary centripetal force for the electron to revolve in its orbit.

$\dfrac{mv^2}{r} = \dfrac{1}{4\pi\epsilon_0}\dfrac{e^2}{r^2}$

Rearranging for the radius $r$:

$r = \dfrac{1}{4\pi\epsilon_0}\dfrac{e^2}{mv^2}$

Substituting the given values:
$\dfrac{1}{4\pi\epsilon_0} = 9 \times 10^9$ N m$^2$ C$^{-2}$
$e = 1.6 \times 10^{-19}$ C
$m = 9 \times 10^{-31}$ kg
$v = \sqrt{25.6} \times 10^5$ m s$^{-1}$

$r = \dfrac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{9 \times 10^{-31} \times (\sqrt{25.6} \times 10^5)^2}$

$r = \dfrac{9 \times 10^9 \times 2.56 \times 10^{-38}}{9 \times 10^{-31} \times 25.6 \times 10^{10}}$

$r = \dfrac{2.56 \times 10^{-29}}{25.6 \times 10^{-21}}$

$r = 0.1 \times 10^{-8}$ m

$r = 1 \times 10^{-9}$ m

Comparing this with $r = x \times 10^{-9}$ m, we get:

$x = 1$

Answer: $1$