The tidal effects related to the difference between gravitional field on 2 opposite points of the surface of the earth. 

As the field is proportional to $\frac{M}{r^2}$, the difference is proportional to its derivative $\frac{M}{r^3}$. (Because the earth diameter is small compared to the distance to moon or sun). 

$$F_s = \frac{M_{sun}}{r_{sun}^3} $$
$$F_m = \frac{M_{moon}}{r_{moon}^3} $$

The relation between that figures is about 0,45. That is: the solar effect on tides of sun is less than half the effect of the moon.