It is inclined to the vertical at an angle $\theta$. The coefficient of friction is $u$? The block can be pushed along the surface only if_______
Clearly, the total normal force on the block will be $$N=Mg+Mg\cos\theta$$Then friction would be $$f=uMg(1+\cos\theta)$$ So $$F\sin\theta \ge f$$ $$Mg\sin\theta \ge uMg(1+\cos\theta)$$ $$\sin\theta \ge u(1+\cos\theta)$$ I can’t go further, the answer is $$\tan \theta /2 \ge u$$
Note that $$\tan\frac{\theta}{2} = \frac{\sin\theta}{1+\cos\theta}.$$ Now, you will be able to push the block if: $$Mg\sin\theta \ge u(Mg+Mg\cos\theta) \Rightarrow \sin\theta \ge u(1+\cos\theta) \Rightarrow \frac{\sin\theta}{1+\cos\theta}=\tan\frac{\theta}{2} \ge u$$