Answer
Please see the work below.
Work Step by Step
We know that
$F=(\frac{GM}{r^2})m$
We plug in the known values to obtain:
$F=(\frac{6.67\times 10^{-11})(4\times 10^6)}{(15)^2})m$
$F=1.1858\times 10^{-6}m$
Now we can find the decrease as
$(\frac{F}{mg})\times 100=(\frac{m(1.1858\times 10^{-6})}{m(9.8)})\times 100=1.2\times 10^{-5}$
This gives the fractional change:
.99999988