Answer
The maximum angle is $4.2^{\circ}$
Work Step by Step
Let's assume that the minimum force required to push the cart up is equal to the component of the cart's weight directed down the ramp:
$mg ~sin(\theta) = 18 ~N$
$sin(\theta) = \frac{18 ~N}{(25 ~kg)(9.80 ~m/s^2)}$
$\theta = sin^{-1}(\frac{18 ~N}{(25 ~kg)(9.80 ~m/s^2)}) = 4.2^{\circ}$
Therefore, the maximum angle is $4.2^{\circ}$.
