$\therefore $ Force required just to move up ${F_{up}} = mg(\sin \theta + \mu \cos \theta )$

Similarly for down ward motion a $ = g(\sin \theta - \mu \cos \theta )$ 

$\therefore $ Force required just to prevent the body sliding down ${F_{dn}} = mg(\sin \theta - \mu \cos \theta )$ 

According to problem ${F_{up}} = 2{F_{dn}}$ 

$⇒$ $mg(\sin \theta + \mu \cos \theta ) = 2mg(\sin \theta - \mu \cos \theta )$ 

$⇒$ $\sin \theta + \mu \;\cos \theta = 2\sin \theta - 2\mu \;\cos \theta $ 

$⇒$ $3\mu \cos \theta = \sin \theta $

$⇒$  $\tan \theta = 3\mu $ 

$⇒$ $\theta = {\tan ^{ - 1}}(3\mu ) = {\tan ^{ - 1}}(3 \times 0.25) = {\tan ^{ - 1}}(0.75)$$ = 36.8^\circ $