Answer
$F=8.99\times10^{-19}N$
Work Step by Step
For identical spherical water drops $q_1=q_2$
Thus from coulomb's law, we have
$F=\frac{q^2}{4\pi{\epsilon_{\circ}}r^2}$
putting the values
$F=\frac{(1.0\times10^{-16})^2}{4(3.14)(8.854\times10^{-12})(1.0\times10^{-2}m)^{2}}$
$F=8.99\times10^{-19}N$
