Let Laurie's annual deposit be L and Kemeny's monthly deposit be K. After a year we have interest of 6.7%, making 1.067*L in savings; we also have after a month 1.00667K in savings. (8% per annum is 0.667% monthly.) After a year we have K((1.00667)^12+(1.00667)^11+...+(1.00667)^2+1.00667) in savings, because the first month's deposit accrues 12 periods of compound interest, the second month's 11 periods, the third month's 10 periods and so on.
This is a geometric progression with common multiplier, r=1.00667, and first term=r, so we can use the expression for the sum of a geometric series S=r(r^n)-1)/(r-1) and the amount after a year (n=12) is 12.533K. After 6 years, then, we have 72 months and n=72, so the total amount is 1.00667(1.00667^72-1)/0.00667K=92.6388K. This must come to $220000, so K=220000/92.6388=$2,374.81 per month.
[Proof of formula for S. S=r+r^2+r^3+...+r^n; rS=r^2+r^3+r^4+...+r^n+r^(n+1); rS-S=S(r-1)=r^(n+1)-r, so S=r(r^n-1)/(r-1).]
For L S=1.067(1.067^6-1)/0.067=7.5751 and the amount accrued after 6 years is 7.5751L, which must come to $220000 and L=220000/7.5751=$29,042.62. Her monthly payment is 29042.62/12=$2,420.22.
So Laurie's monthly payment is $2,420.22 and Kemeny's is $2,374.81.
The difference in interest is given by 220000-72K-(220000-6L)=6L-72K=6*29042.62-72*2374.81=$3,269.40.