According to the principle of displacement, when an object is immersed in liquid, it displaces an equal volume of liquid and according to Archimedes' principle, the upward buoyant force that is exerted on a body immersed in a fluid, whether partially or fully submerged, is equal to the weight of the fluid that the body displaces and acts in the upward direction at the centre of mass of the displaced fluid.
 

Density of block =5 gm/cm²
Volume of block V=(5×5×5)cm³ =125cm³

Mass of block M=density × volume =5×125 =0625 kg

Now as the block is stationery upwards and downward form on the block should balance each other.

$\therefore$N + Mg = F_B

$\therefore$Buoyant force, F_B​= density of water × volume of block × g

=(1 gm/cm³)×(125cm³)×(98 m/s²)
=(125 gm)×(98 m/s²)
=(0.125×9.8) kgm/s²
=(0.125×9.8)N(1 N=1 kgm/s²)
Putting values is (1)

N=(0.625×9.8)−(0.125×9.8) Newton

⇒ Apparent weight =(0.5×9.8) Newton

(∵ Apparent weight = Normal force)

Now, 1 gf(gram−force)=9.8×10⁻³N

⇒1 N=9.81​×10³ gf

Thus, Apparent weight +(0.5×9.8)×(9.81​×103)

=500 gf

Hence apparent weight of the block completely submerged in water is 500 gram- force