a
As given in fig.
Normal force,
$N=(10-4) \hat{j}=6 \hat{j}$
Maximum frictional force,
$f=\mu N(\hat{-} i)=0.3 \times 10(\hat{-} i)=3(-\hat{i})$
Maximum frictional forceis $f>1 N,$ so that the frictional force act only equal to the force in $(+\hat i) $direction but in opposite direction
$f=1(-\hat{i})$