Answer
$0.85N$
Work Step by Step
We know that
$\Sigma F_y=Ncos30^{\circ}+Ncos30^{\circ}-mg=0$
This can be rearranged as:
$N=\frac{mg}{2cos30^{\circ}}$
We plug in the known values to obtain:
$N=\frac{(0.15)(9.81)}{2cos30^{\circ}}$
$N=0.85N$