Answer
3.1 N
Work Step by Step
$N=mg-P=(2.5\,kg\times9.8\,m/s^{2})-12\,N=12.5\,N$
$f_{s_{max}}=\mu_{s}N=0.40\times12.5\,N=5.0\,N$
Applied force (6.0 N) exceeds $f_{s_{max}}$ and therefore the block starts to move. In this case, the force of kinetic friction is
$f_{k}=\mu_{k}N=0.25\times12.5\,N=3.1\,N$