Chemistry help here, please

Hello there! Thanks for your help in advance. I have a question I'm pretty at a loss on whilst doing homework. If you can please held, I'd really appreciate it!

Here it is: I'm given 50.0 g cube, then the temperature of water in a beaker (20degreesC). Then the cube is put inside of the beaker with water and measures 5.6 g. I am given the d=0.99821 g/cm^3.

The question then asks to give the density of the block in g/cm^3.

Thanks guys and gals!

{m (on scale) x g} + {ρ x V x g} - (m (block) x g} = 0

Dividing both sides of equation by g:
m (on scale) + (ρ x V) - m (block) = 0 ~(eq2)

Given: V = volume of water displaced
= 40% volume of the block = 0.40 x V (block)

Substituting 0.40 x V (block) into ~(eq2):
m (on scale) + (ρ x 0.40 x V (block)) - m (block) = 0 ~(eq3)
where,
m (on scale) = 5.6 g
ρ = density of water at 20ºC = 0.99821 g/cm^3
m (block) = 50 g

Substituting m (on scale), ρ and m (block) into ~(eq3):
5.6 g + (0.99821 g/cm^3 x 0.40 x V (block)) - 50 g = 0 ~(eq4)

Simplifying ~(eq4) and solving for V (block):
==> 5.6 g + 0.399284 g/cm^3 x V (block) - 50 g = 0
==> 0.399284 g/cm^3 x V (block) - 44.4 g = 0
==> 0.399284 g/cm^3 x V (block) = 44.4 g
==> V (block) = 44.4 g / (0.399284 g/cm^3)
==> V (block) = 111.199 cm^3

ρ (block) = density of block = m (block) / V (block)
= 50 g / 111.199 cm^3
= 0.449644 g / cm^3

*** Answer: 0.44964 g / cm^3 ***

(at 5 decimal places, like density of water)