Make a sketch. I will assume that you just did it and look in it.


The radius of the wheel is R = 300 mm 0.3 m.


The weight of 5 kN is applied at the center of the wheel and directed vertically down.


The force  F  (under the question) is applied horizontally at the center of the wheel. 

As I said, it is directed horizontally, and it is directed to the block's side.


Draw the radius OA of the wheel from its center O to the point A where the wheel contacts with the block.


Notice that the angle the radius OA makes with the horizontal line is  = 30°.
It is because the radius of the wheel is 300 mm,  while the height of the block is 150 mm.


The weight produces the moment of force  =  W*h,  where the arm "h" is the horizontal projection of the radius OA

    h =  = R*cos(30°).

Therefore the moment of force produced by the wheel's weight is  W*h =  Newtons*meter, 
 where 0.3 = 0.3 m is the radius of the wheel.


The force F produces the moment   = F*0.15 Newtons*meter.


As you understand (or should understand), the moments of forces, produced by W and F, have opposite directions, and
the equation for the least value of the horizontal force F, through the center of the wheel, required just to turn the wheel 
over the corner A of a block is the equality of magnitudes of the two moments of forces  (it is the key moment of the solution !)


    0.15*F = ,   or

    0.15*F = ,   which implies


    F =  =  Newtons.


Answer.  The least value of the horizontal force F, through the center of the wheel, required just to turn the wheel 
over the corner A of a block is  Newtons = 8660 Newtons  (approximately).


Regarding the reaction force from the block to the wheel,  it is directed along the radius OA from A to O and is equal to

     =  =  =  = 10 kilo-Newtons.

    Those individuals who solved at the level  Math+++, were eligible to be enrolled without the entrance exams.

    Later, as a rule, they became professors at universities and some of them became Fields' award laureates in Math . . .