As shown in the figure, a metal piece having a mass of 160 g rests...

Given:
Mass of the metal piece, $m = 160\text{ g} = 0.16\text{ kg}$
Density of the metal, $\rho_m = 8000\text{ kg m}^{-3}$
Density of water, $\rho_w = 1000\text{ kg m}^{-3}$
Acceleration due to gravity, $g = 10\text{ m s}^{-2}$

The volume of the metal piece is:
$V = \dfrac{m}{\rho_m} = \dfrac{0.16}{8000} = 2 \times 10^{-5}\text{ m}^3$

Since the metal piece makes contact with the bottom at only a few points, water is present under it, and the buoyant force acts on its entire volume.
The buoyant force $F_B$ is:
$F_B = V \rho_w g = (2 \times 10^{-5}) \times 1000 \times 10 = 0.2\text{ N}$

The weight of the metal piece $W$ is:
$W = mg = 0.16 \times 10 = 1.6\text{ N}$

For the metal piece in equilibrium, the upward forces balance the downward forces:
$N + F_B = W$
$\Rightarrow N = W - F_B$
$\Rightarrow N = 1.6 - 0.2 = 1.4\text{ N}$