Answer
$1.6 \times 10^8\;\rm N/C$
Work Step by Step
We are given the difference between the electric potential energy in each case of the water dipole molecule where
$$\Delta U_{\rm dipole}=U_{\rm perpendicular}-U_{\rm aligned}=\bf 1\times 10^{-21}\rm \;J$$
And we know that in one case, it is perpendicular to our electric field while in the case other it is aligned.
We know that the electric potential energy of a dipole is given by
$$U_{\rm dipole}=-pE\cos\phi$$
Hence,
$$ U_{\rm perpendicular}-U_{\rm aligned} = -pE\cos90^\circ-(-pE\cos0^\circ) $$
Recall that $\cos90^\circ=0$ and $\cos0^\circ=1$
$$ U_{\rm perpendicular}-U_{\rm aligned} =-0+pE $$
Therefore,
$$E=\dfrac{ U_{\rm perpendicular}-U_{\rm aligned} }{p}$$
Plug the known;
$$E=\dfrac{ 1\times 10^{-21}}{6.2\times 10^{-30}}$$
$$E=\color{red}{\bf 1.6 \times 10^8}\;\rm N/C$$
