Dissociation constant of water at $25°\mathrm{C}$ is

It is given to find Dissociation constant of water at $25°\mathrm{C}$
Pure water contains ions. There is self-ionisation constant of water, ${\mathrm{K}}_{\mathrm{w}}$, which has no units. It is the product of the hydronium ions times the hydroxide ions,
${\mathrm{K}}_{\mathrm{w}}$=$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]$
the ions are equal in concentrations, which experiments have shown to be
$1.0\times {10}^{-7}$for each one at $25°\mathrm{C}$
$\Rightarrow {\mathrm{K}}_{\mathrm{w}}$=$\left({10}^{-7}\right)\left({10}^{-7}\right)$
${\mathrm{K}}_{\mathrm{w}}$=$1.0\times {10}^{-14}$
$\Rightarrow$ At $25°\mathrm{C}$, the dissociation constant for pure water is given by $1.0\times {10}^{-14}$.
Hence the correct answer (A) $1.0\times {10}^{-14}$