$\mu =\sqrt {\mu_1^2 +\mu_1^2 +2 \mu_1 \mu_1 \cos \alpha}$
In $H_2O$ only two dipoles equal to $\mu_1$ are operating due to two $O-H$ bonds.
Thus, $1.84 =\sqrt {(1.5)^2+(1.5)^2+2 (1.5 )\times (1.5) \cos \alpha}$
$\cos \alpha=-0.2476$
$\alpha= 140^{\circ}20'$
Hence C is the correct answer.