First off, we can figure out how many mols of sulfuric acid were used to titrate the potassium hydroxide solution.  To do that we take the molarity (mol/L) and multiply it by the volume used in the titration (in liters) which will cancel the units of volume and leave us with the number of mols that were consumed in the titration:

.0142 L x 1.50 (mol/L) = 0.0213 mol

Now, this is the number of mols of sulfuric acid that were used to neutralize the potassium hydroxide, we know from the above equation that for each mol of sulfuric acid there were two mols of potassium hydroxide, so if we multiply the above number by two we get the number of mols of potassium hydroxide in the solution:

0.0213 mol H₂SO₄ x 2 (mol KOH/mol H₂SO₄) = 0.0426 mol KOH

Now we finally know exactly how many mols of potassium hydroxide there were in that 90 ml sample.  Remember here that molarity is just mol/L and we've got both of those measurements:

0.0426 mol KOH / 0.09 L = 0.473 M

So, the molarity of the potassium hydroxide solution given is 0.473 mol/L based on the amount of known concentration sulfuric acid solution used in the titration.