Answer
See answers.
Work Step by Step
a. The 2 objects have the same apparent weight when submerged. However, under these conditions, the object with the larger volume gets a bigger upward boost from the buoyant force, because it displaces more water. Therefore the larger object actually weighs more in air.
b. We have the following condition when the apparent weights are equal. Let L be the larger object, and S the smaller one.
$$W_L-\rho_{water}V_Lg = W_S-\rho_{water}V_Sg $$
$$\frac{W_L-\rho_{water}V_Lg}{V_L} \lt \frac{W_S-\rho_{water}V_Sg }{V_S}$$
$$\frac{m_Lg}{V_L}-\rho_{water} g \lt \frac{m_Sg}{V_S}-\rho_{water} g $$
$$\frac{m_Lg}{V_L} \lt \frac{m_Sg}{V_S} $$
$$\rho_L g \lt \rho _S g $$
$$\rho_L \lt \rho _S $$
The smaller object is denser.