Answer : 0.54

The component of the weight down the incline $=8(9.8) (0.5) =39.2 \;N$

Now $F=ma$ leads to $39.2 -f=8(0.3)$, So that $f= 36.8 \;N$ .

The Normal force equals the component of W perpendicular to the incline , $8(9.8)(0.867)=67.9\;N$.

Therefore $\mu =36.8/67.9 =0.54$