Answer
The net force on the Moon points toward the Sun.
Work Step by Step
We know that the force on the Moon by the Sun is given as
$F_{M.S}=G\frac{M_M \space M_S}{R_{M.S}^2}$
We plug in the known values to obtain:
$F_{M.S}=6.67\times 10^{-11}\times \frac{7.35\times 10^{22}(2.0\times 10^{30})}{(1.50\times 10^{11})^2}$
$F_{M.S}=4.35\times 10^{20}N$
The force on the Moon by the Earth is given as
$F_{M.E}=G\frac{M_M \space M_E}{R_{M.E}^2}$
We plug in the known values to obtain:
$F_{M.S}=6.67\times 10^{-11}\times \frac{7.35\times 10^{22}(5.97\times 10^{24})}{(3.84\times 10^{8})^2}$
$F_{M.S}=1.98\times 10^{20}N$
We can see that the Sun exerts more force on the Moon as compared to the Earth and thus the net force on the Moon points toward the Sun.
