Answer
$\rho_{Ar}=1.78\ g/L$
Work Step by Step
The problem states that the air density is the sum of the products of densities of its constituents by their volume fraction, so from the data given:
Air (dry, $CO_2$ free): $\rho= 1.29327 \ g/L$, $20.96\%\ O_2,\ 78.11\%\ N_2,\ 0.930\%\ Ar$
$O_2$:$\rho= 1.42952\ g/L$
$N_2$:$\rho= 1.25092\ g/L$
So:
$1.29327=20.96/100\times1.42952+78.11/100\times1.25092+0.930/100\times \rho_{Ar}$
$\rho_{Ar}=1.78\ g/L$