An acid solution of pH = 6 is diluted 1000 times, the pH of the final solution  is

pH = 6 means $\left[{\mathrm{H}}^{+}\right] = {10}^{-6} \mathrm{M}$. After dilution, the hydrogen ion concentration becomes ${10}^{-9} \mathrm{M}$. Under such conditions, the hydrogen ions obtained from water cannot be neglected.

$\therefore \mathrm{Total} \left[{\mathrm{H}}^{+}\right] = {10}^{-9} + {10}^{-7} \left(\mathrm{approx}.\right)$

$= {10}^{-7} \left({10}^{-2} + 1\right) = {10}^{-7} \left(1.01\right)$

[The contribution of ${\mathrm{H}}^{+}$ from water will not be exactly ${10}^{-7}$ but still we can make an approximation in an objective problem and take it as ${10}^{-7}$]

$\mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right] =-1.01 \times {10}^{-7} =7 - 0.0043 = 6.9957$