She should invest $6491.73.
The equation we use to solve this is in the form

,
where A is the total amount in the account, p is the principal invested, r is the interest rate as a decimal, n is the number of times per year the interest is compounded, and t is the amount of time.
A in our problem is 14000.
p is unknown.
r is 6% = 6/100 = 0.06.
n is 2, since it is compounded semiannually.
t is 13.
