When the solution is very dilute, the concentration of $ OH^{-} $ produced from water cannot be neglected. Hence,
$\left[O H^{-}\right]=10^{-10}+10^{-7}$
(obtained from water)
$=10^{-7}(0.001+1)$
$=1.001 \times 10^{-7} $
$\therefore p O H=-\log \left[O H^{-}\right]$
$=-\log \left(1.001 \times 10^{-7}\right) $
$=7-0.01=6.99$
$\therefore p H=14-p O H$
$=14-6.99=7.01$