To start, let's write the chemical equation for this reaction:

#"HNO"_3 "(aq)" + "LiOH (aq)" rarr "LiNO"_3 "(aq)" + "H"_2"O (l)"#

Let's find the number of moles of #"LiOH"# reacting using the molarity equation.

#"mol LiOH" = (0.100 "mol"/cancel("L"))(0.03690cancel( "L")) = 0.00369 "mol LiOH"#

For this calculation, you would need to have converted from #"mL"# to #"L"#, I just did not show it here (I'm assuming you know how to).

Since all the coefficients in the chemical equation are equal (#1#), the relative number of moles of that react is also #0.00369 "mol"#.

We now know the moles of #"HNO"_3# and the volume of #"HNO"_3# that reacted, so by using the molarity equation again, the molarity of the nitric acid solution is

#M = (0.00369 "mol")/(0.0500 "L") = color(red)(0.0738 "mol"/"L"#