A photon of mass $m$ and frequency v falls through a height $H$ through the earth's gravitational field as shown in figure below. 

When the photon falls through the earth's gravitational field, then it is gains extra energy. Therefore,

Final photon energy $=$ Initial photon energy $+$ Increase in energy 

$\Rightarrow h{v}^{\text{'}}=hv+mgH$

$=hv+\frac{hv}{c^2}.gH=hv\left(1+\frac{gH}{c^2}\right)$

So, $\frac{f^{\text{'}}-f}{f}=$ fractional change in frequency $=\frac{gH}{c^2}=\frac{10\times 1000}{{\left(3\times {10}^8\right)}^2}=\frac{10}{9}\times {10}^{-13}$.