When two solid pieces of steel and aluminium are immersed completely in water, their weights are equal. When weighted in air, the solid pieces:

The upward buoyant force exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to the weight of the fluid that the body displaces, according to Archimedes' principle.

When an object is totally or partially submerged in fluid, an upward force is imposed on the body. The buoyant force is the name given to this force.

Given that when immersed completely in water, both have equal weights

$\mathrm{Mass} \left(\mathrm{m}\right) =\mathrm{density} \left(\mathrm{d}\right)\times \mathrm{Volume} \left(\mathrm{V}\right)$

since Density of steel > density of aluminum

So ${\mathrm{V}}_{\mathrm{Steel}}<{\mathrm{V}}_{\mathrm{Al}}$

$$\begin{gathered} {\mathrm{W}}_{\mathrm{steel}}={\mathrm{W}}_{\mathrm{Steel}}-{\mathrm{\rho }}_{\mathrm{f}}{\mathrm{V}}_{\mathrm{steel}} \mathrm{g} \\ {\mathrm{W}}_{\mathrm{Al}\text{'}}={\mathrm{W}}_{\mathrm{Al}}-{\mathrm{\rho }}_{\mathrm{f}}{\mathrm{V}}_{\mathrm{Al}} \\ \therefore {\mathrm{W}}_{\mathrm{Steel}}-{\mathrm{\rho }}_{\mathrm{f}}{\mathrm{V}}_{\mathrm{Steel}} \mathrm{g}={\mathrm{W}}_{\mathrm{Al}}-{\mathrm{\rho }}_{\mathrm{f}}{\mathrm{V}}_{\mathrm{Al}} \mathrm{g} \end{gathered}$$

Since ,

$$\begin{gathered} {\mathrm{V}}_{\mathrm{Steel}}<{\mathrm{V}}_{\mathrm{Al}} \\ \mathrm{So}, {\mathrm{W}}_{\mathrm{Steel}} > {\mathrm{W}}_{\mathrm{Al}} \end{gathered}$$

So the correct answer is option A.