Answer
$F=122.5N$ uphill parallel to the inclined surface.
Work Step by Step
The component of the weight parallel to the inclined surface is responsible for providing acceleration to the block i.e. $wsin\theta=mgsin\theta$
Smallest such force applied so that the body does not accelerate is $F=mgsin\theta=25\times9.8sin30^{\circ}=122.5N$ uphill parallel to the inclined surface.
