1) It is missing the amount (mililiters) of the 0.045 molar barium hydroxide.
2) I will show you how to solve this using a generic value V, for this volumen. And, then I will calculate the requested molarity for some assumed values of V.
3) Start with stating the balanced chemical equation:
Ba(OH)₂ + 2HNO₃ → Ba(NO₃)₂ + 2H₂O
4) State the mole ratio between the two reactants:
1 mol Ba(OH)₂ : 2 mol HNO₃
5) Set the equations for the number of moles of both solutions
M = n / V ⇒ n = M × V
⇒ number of moles of baryum hidroxide: 0.045V
number of moles of nitric acid: 0.0385 M
6) As per the theoretical mole ratio, the neutralization implies number of moles of barium hydroxide equals 2 times the number of moles of nitric acid:
⇒ 0.045V = 2×0.0385×M
And, solve for M: M = 0.045V / 0.077 = 0.5844V
That is the fomula that you can use: M = 0.5844V.
7) Here some examples, for different values of the volume V.
Volum in ml V (in liters) M = 0.5844V
10 10/1000 0.0058
20 20/1000 0.012
30 30/1000 0.018
38.5 38.5 / 1000 0.023
77 77/1000 0.045