What volume of vector and insert are needed to get a 1:1 molar ratio in 20 micro-gram ligation reaction?

You want to clone a 200 bp (base pair) insert into a 3200 bp vector. The concentrations of the starting stocks of insert and vector 1 micro-gram per micro-liter. You want the following amounts of vector and insert in a 20 micro-gram ligation reaction:

-total of 5 micro-gram of vector

-1:1 molar ratio of vector to insert

What volume of your 1 micro-gram per micro-liter vector and what volume of your 1 micro-gram per micro-liter insert you need to add? (1 mole of base pairs = 660g)

I've got an idea and I just need to check if it is correct. For both vector and insert I have found out that there is 15.15 x10^(-10) moles per micro-gram.