Answer
1.7 million volts.
Work Step by Step
The potential of the Earth goes up because the previously electrically neutral Earth will now be charged. The new charge will be the elementary charge, e, multiplied by the number of electrons removed.
$$Q=(1.602\times10^{-19}C)(\frac{10e}{H_2O)})(\frac{6.02\times10^{23}H_2O}{0.018kg})(\frac{1000kg}{m^3})\frac{4\pi}{3}(0.00175m)^3=1201C$$
Assume that the excess charge is distributed symmetrically.
$$V=\frac{kQ}{R_{Earth}}=\frac{8.99\times10^9}{6.38\times10^6 m}(1201C) =1.7\times10^{6}V $$
