$$
\begin{aligned}
& F_1=m g \sin \theta+\mu m g \cos \theta \\
& F_2=m g \sin \theta-\mu m g \cos \theta
\end{aligned}
$$
Given that, $F_1=3 F_2$
or $\left(\sin 45^{\circ}+\mu \cos 45^{\circ}\right)$
$$
=3\left(\sin 45^{\circ}-\mu \cos 45^{\circ}\right)
$$
On solving, we get $\mu=0.5$
$$
\therefore \quad N=10 \mu=5
$$
$\therefore$ Answer is 5 .
Analysis of Question
Question is simple. Only one has to take care of direction and magnitude of friction.