Missing detail:
The coefficient of static friction is not specified in the problem. Searching online, I found that for wood-on-wood contact it is between 0.25-0.5. I assume here the maximum value, 0.5.
Solution:
When pushing the crate horizontally with a force F, only two forces act horizontally on the crate: F and [tex]F_f[/tex], the frictional force, which points into the opposite direction of F and whose magnitude is given by
[tex]F_f = \mu_S m g[/tex]
where [tex]\mu_S=0.5[/tex] is the coefficient of static friction, [tex]m=150 kg[/tex] is the mass of the crate and [tex]g=9.81 m/s^2[/tex] is the gravitational acceleration.
In order to start to move the crate, the force F applied must overcome the frictional force. This means that the minimum value of F necessary is equal to Ff:
[tex]F_{min}=F_f = \mu_S mg=(0.5)(150 kg)(9.81 m/s^2)=735.7 N[/tex]
