Answer
6.0 N
Work Step by Step
$N=W-P= mg-P$$=(2.5\,kg\times9.8\,m/s^{2})-8.0\,N=16.5\,N$
$f_{s_{max}}=\mu_{s}N=0.40\times16.5\,N=6.6\,N$
Applied force which is equal to $6.0\,N$ doesn't exceed the maximum static friction ($f_{s_{max}}$). This implies that the block will not slide. In that case, frictional force is equal in magnitude to the applied force. Therefore, magnitude of the frictional force acting on the block is $6.0\,N$.
