The reaction for HNO3 ad KOH is given by,

HNO3 + KOH -------> KNO3 + H2O

Therefore 1 mol of HNO3 will neutralize 1 mol of KOH.

The amount of KOH in 39 mL of 2.0 M KOH is,

                        = (2/1000) x 39 mol

                        = 0.078 mol

Therefore mols of HNO3 needed is = 0.078 mol

Volume of 6.0 M HNO3 needed is, v then,

0.078 = (6/1000) x v

v = 13 mL.

Therefore, the volume of 6.0 M HNO3 needed to neutralize 39mL of 2.0 KOH is 13 mL.