To solve this problem it is necessary to resort to the concepts expressed in the Buoyancy Force.
The buoyancy force is given by the equation

Where,
Density
V =Volume
g = Gravitational Acceleration
PART A) From the given data we can find the volume, so


PART B) The mass can be expressed from the Newton equation in which

Where,
m = mass
g = Gravitational acceleration
Replacing with our values we have that


Therefore the Density can be calculated with the ratio between the Volume and Mass



Therefore the Density of the log is
