#### Answer

(a) The weight of the block is $0.794~N$
(b) The scale reading is $0.544~N$

#### Work Step by Step

(a) We can find the mass $M$ of the block:
$M = \rho~V$
$M = (2700~kg/m^3)(0.020~m)(0.030~m)(0.050~m)$
$M = 0.081~kg$
We can find the weight of the block:
$weight = Mg = (0.081~kg)(9.80~m/s^2) = 0.794~N$
The weight of the block is $0.794~N$
(b) According to Archimedes' principle, the buoyant force is equal to the weight of the displaced oil. We can find the buoyant force:
$F_b = m_o~g$
$F_b = \rho_o~V~g$
$F_b = (850~kg/m^3)(3.0\times 10^{-5}~m^3)(9.80~m/s^2)$
$F_b = 0.250~N$
We can find the scale reading when the block is submerged in oil:
$F_{scale} = 0.794~N-0.250~N = 0.544~N$
The scale reading is $0.544~N$