The answer is 3.0 "g" of Ba(SO_4) will form in this reaction.
Your general chemical reaction is
Na_2SO_(4(aq)) + Ba(NO_3)_(2(aq)) -> BaSO_4(s) + 2NaNO_(3(aq))
Notice that you've got a 1:1 mole ratio between Ba(NO_3)_2 and BaSO_4; this means that for every mole of Ba(NO_3)_2 used, 1 mole of solid will be produced.
The number of Ba(NO_3)_2 moles can be determined using its molarity:
C=n/V => n_(Ba(NO_3)_2) = C * V = 0.50 "M" * 25*10^(-3) "L"
n_(Ba(NO_3)_2) = 0.013 "moles"
This means of course that n_(BaSO_4) = n_(Ba(NO_3)_2) = 0.013 moles of solid will be formed. Knowing that BaSO_4's molar mass is 233.3 "g/mol", the mass produced will be
m_(BaSO_4) = n_(BaSO_4) * 233.3 g/(mol) = 0.013 "moles" * 233.3g/(mol)
m_(BaSO_4) = 3.0 "g"
The reaction's complete ionic equation is
2Na_((aq))^(+) + SO_(4(aq))^(2-) + Ba_((aq))^(2+) + 2NO_(3(aq))^(-) -> BaSO_(4(s)) + 2Na_((aq))^(+) + 2NO_(3(aq))^(-)
The net ionic equation is
Ba_((aq))^(2+) + SO_(4(aq))^(2-) -> BaSO_(4(s))
