Well to get you started out this is a problem where you're going to have to neutralize the moles of NaOH you have with an equal number of moles of HNO3. Your molar weight for NaOH is roughly 40 g/mol and rember that molarity = moles per liter, meaning that for .0250 M HNO3 there's .0250 moles of acid per liter of solution you add.
17.5/40 = ?
? = ??(.0250)
Good luck.