The $\mathrm{H}{ }^{+}$ion concentration of an aqueous solution is $0.001\mathrm{M}$ or $1\times {10}^{-3}\mathrm{M}$

Since.  we know that $\mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right]$

Using this equation, by plugging in the values

$\mathrm{pH}=-\log {10}^{-3}=-(-3) \log 10=3$

$\mathrm{pH}=3$

We know that

$\mathrm{pOH}=14-\mathrm{pH}=14-3=11$

$\therefore \mathrm{pOH}=11$