Answer
$23.1\;\rm N$
Work Step by Step
Since the box slides down, so the friction force must be up.
See the force diagram of the box below.
Since we need the box to move down at a constant velocity, the net force exerted on it must be zero.
$$\sum F_x=F_n-F_{push}\cos45^\circ=ma_x=m(0)=0$$
Thus,
$$F_n=F_{push}\cos45^\circ\tag 1$$
$$\sum F_y=f_k+F_{push}\sin45^\circ-mg=ma_y=m(0)=0$$
Hence,
$$f_k+F_{push}\sin45^\circ-mg=0$$
$$\mu_kF_n+F_{push}\sin45^\circ =mg $$
Plugging from (1);
$$\mu_kF_{push}\cos45^\circ+F_{push}\sin45^\circ=mg $$
$$F_{push}\left[\mu_k\cos45^\circ+ \sin45^\circ\right]=mg $$
$$F_{push}=\dfrac{mg}{\left[\mu_k\cos45^\circ+ \sin45^\circ\right]} =\dfrac{2\cdot 9.8}{\left[0.2\cos45^\circ+ \sin45^\circ\right]} $$
$$F_{push}=\color{magenta}{\bf23.1}\;\rm N$$
