$\begin{array}{1 1} Mg\;\sin \theta \\ Mg\;cos \theta \\ 2\;Mg\;\cos \theta \\ 2\;Mg\;\sin \theta \end{array} $

Answer : $ 2\;Mg\;\sin \theta$

The component of weight $Mg$ of the block along the inclined plane $=Mg \sin \theta$.

The minimum frictional force to be overcome is also $Mg \sin \theta$

To make the block just move up the plane the minimum force applied must overcome the component $Mg \sin \theta$ of gravitational force as well as the frictional force $Mg \sin \theta = 2 \;Mg \sin \theta$.

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