The answer is #2.79g#.
Starting from the balanced chemical equation
#Ba(NO_3)_(2(aq)) + Na_2SO_(4(aq)) -> BaSO_(4(s)) + 2NaNO_(3(aq))#
Taking into consideration the solubility rules (more here: http://www.chem.sc.edu/faculty/morgan/resources/solubility/), we can see that #Ba(NO_3)_2#, #Na_2SO_4#, and #NaNO_3# will dissociate into their respective ions, which will lead to the reaction's net ionic equation
#Ba_((aq))^(2+) + SO_(4(aq))^(2-) -> BaSO_(4(s))#
Since sulfate compounds formed with #Ba^(2+)# cations are insoluble in water (only slightly soluble, #BaSO_4#'s #K_(sp)# being equal to #1.1 * 10^(-10)# ), this double-replacement reaction forms a precipitate, #BaSO_4#.
We know from the balanced chemical equation that the mole-to-mole ratio of #Ba(NO_3)_2# and #BaSO_4# is 1:1; that is, for every mole of barium nitrate used, one mole of barium sulfate is produced.
The number of barium nitrate moles can be determined from its molarity, #C = n/V#
#n_(Ba(NO_3)_2) = C * V = 0.500 M * 25.0 * 10^(-3) L = 0.012#
Knowing barium sulfate's molar mass (#233.3 g/(mol)#), and the number of moles produced, we get
#m_(BaSO_4) = n_(BaSO_4) * molarmass = 0.012 mol es * 233.3g/(mol e) = 2.79g#
