$V=1000$ dm$^3$ = 1 m$^3$. Thus, given the density of air $d = 1.225$ kg/m$^3$, the gravity $g=9.81$ m/s$^2$, the force exercised by air over the object should be $F = dV g = 12.0$ N.

Do you agree?

This is an undergrad physics problem, which puzzled me because of the little background provided; see also the comment below by @RC_23.

Problem. Compute the force exercised by the air to an object whose volume is $V=1000$ dm$^3$.

Solution. Now $V=1000$ dm$^3$ = 1 m$^3$. My reasoning is that the object is pressing the air causing a shift in its mass. This shift must be equal to the hydrostatic force, which, assuming the object is in equilibrium, must be equal to the gravity force.

Thus, given the density of air $d = 1.225$ kg/m$^3$ and the gravity $g=9.81$ m/s$^2$, the force exercised by air over the object should be $$F_p = mg = (dV) g = 12.0\,\mathrm{N}.$$

                                                     [solution: 12.7 N]

Do you agree with my reasoning? If not, how would you attack the problem?

$V=1000$ dm$^2$ = 1 m$^2$. Thus, given the density of air $d = 1.225$ kg/m$^3$, the gravity constant $g=9.81$ m/s$^2$, the force exercised by air over the object should be $F = dV g = 12.0$ N.

Do you agree?

$V=1000$ dm$^3$ = 1 m$^3$. Thus, given the density of air $d = 1.225$ kg/m$^3$, the gravity $g=9.81$ m/s$^2$, the force exercised by air over the object should be $F = dV g = 12.0$ N.

Do you agree?