Start by taking a look at the balanced chemical equation for this double replacement reaction
#Na_2CrO_(4(aq)) + Sr(NO_3)_(2(aq)) -> SrCrO_text(4(s]) + 2NaNO_(3(aq))#
Notice that you have #1:1# mole ratios between the reactants, sodium chromate and and strontium nitrate, and between the ractants and the precipitate, strontium chromate.
This means that your reaction will consume equal numbers of moles of reactants and produce the same number of moles of strontium chromate.
Use the volumes and molarities of the two solutions to figure out how many moles of each reactant you're mixing together
#C = n/V => n = C * V#
#n_(Na_2CrO_4) = "0.40 M" * 100.0 * 10^(-3)"L" = "0.040 moles"# #Na_2CrO_4#
#n_(Sr(NO_3)_2) = "0.30 M" * 150.0 * 10^(-3)"L" = "0.045 moles"# #Sr(NO_3)_2#
Notice that you have more moles of strontium nitrate than you have of sodium chromate, which means that the latter will act as a limiting reagent, i.e. it will limit the amount of sodium chromate that will react.
The #1:1# mole ratio will consume all the moles of sodium chromate and leave you with an excess of
#n_(Sr(NO_3)_2) = 0.045 - 0.040 = "0.0050 moles"# #Sr(NO_3)_2#
This means that the reaction will produce
#0.040color(red)(cancel(color(black)("moles"Na_2CrO_4))) * ("1 mole "SrCrO_4)/(1color(red)(cancel(color(black)("mole"Na_2CrO_4)))) = "0.0040 moles"# #SrCrO_4#
To determine the mass of precipitate produced by the reaction, use strontium chromate's molar mass
#0.040color(red)(cancel(color(black)("moles"))) * "203.6 g"/(1color(red)(cancel(color(black)("mole")))) = "8.144 g"#
Rounded to two sig figs, the number of sig figs you gave for the molarities of the two solutions, the answer will be
#m_(SrCrO_4) = color(green)("8.0 g")#