Answer
(a) $750\frac{Kg}{m^3}$
(b) increases
Work Step by Step
(a) We know that
$\rho_{log}=\rho_w{\frac{V_{submerged}}{V}}$
We plug in the known values to obtain:
$\rho=(1000)(\frac{3}{4})=750\frac{Kg}{m^3}$
(b) The portion of the log carried increases because the density of the displaced fluid is directly proportional to the bouyant force.