Here's what I got.
The idea here is that the atomic mass that is listed in the Periodic Table is actually an average atomic mass because it is calculated by using the atomic masses of the stable isotopes of a given element and their respective abundances.
More specifically, each isotope will contribute to the average atomic mass of the element proportionally to its abundance.
For an element that has
#"avg atomic mass" = sum_i "atomic mass isotope"_i xx "decimal abundance"_i#
Now, you know that chlorine has two stable isotopes, chlorine-35 and chlorine-37.
Since the problem doesn't provide you with the exact atomic masses of the two isotopes, you can say that you have
#"For chlorine-35: " "atomic mass " ~~ " 35 u"#
#"For chlorine-37: " "atomic mass " ~~ " 37 u"#
Since there are only two stable isotopes, you can say that their percent abundances must add up to give
If you take
This means that you have
#35.453 color(red)(cancel(color(black)("u"))) = overbrace(35 color(red)(cancel(color(black)("u"))) * x)^(color(blue)("the contribution of chlorine-35")) + overbrace(37 color(red)(cancel(color(black)("u"))) * (1-x))^(color(blue)("the contribution of chlorine-37"))#
Rearrange to find the value of
#35.453 = 35x + 37 - 37x#
#2z = 37 - 35.452 implies x = (37 - 35.453)/2 = 0.7735#
Consequently, you will have
#y = 1 - 0.7735 = 0.2265#
This means that the percent abundances of the two isotopes will be equal to
#"For chlorine-35: " 77.35%#
#"For chlorine-37: " 22.65%#
I'll leave the two values rounded to four sig figs.