I am assuming this is a 1,000g mixture.

Question

42% H3BO3 = 420g H3BO3/(420g H3BO3+580g H2O)= 0.42 by mass.

420g H3BO3/(420g H3BO3+580g H2O+Xg H2O)= 0.06
Solve for X
Check my thinking. Check my work.

There is an alternative approach. You want to dilute it by a factor of seven. (42 to 6). So add six parts water.

Six liters of water added to the one liter of boric acid. It gives you the same answer.

Caleb has 1 liter of a mixture containing 42% boric acid. How much water must be added to make the mixture 6% boric acid?

I don't get how to do weighted averages