I’ll work in kg-equivalent-masses first, converting to N (Newtons) only at the end.
The kg-equivalent forces are those of the weight due to a mass of 2.0 kg acting downwards (short “perpendicular side” of triangle), of
2 sqrt(3) kg horizontally (long perp. side) and of 4.0 kg along the hypoteneuse, normal to the board.
(The ratios of these sides, 1sqrt(3)2 are the ratios for the 30, 60, 90 deg. triangle.)
So the normal force exerted by the book is the force due to the weight of a 4.0 kg mass, that is
4.0 x 9.8 N =
And the answer is
Have a nice day