The pH of a solution is 6.72 ± 0.03. What is the concentration of H in the solution and its absolute error? [H+]= __M ± ___M?

Average [H⁺] = 10^⁻⁶·⁷² M = 1.91 × 10⁻⁷ M

When pH = 6.72 + 0.03 = 6.75 :
min. [H⁺] = 10^⁻⁶·⁷⁵ M = 1.78 × 10⁻⁷ M

When pH = 6.72 - 0.03 = 6.69 :
max. [H⁺] = 10^⁻⁶·⁶⁹ M = 2.04 × 10⁻⁷ M

(1.91 × 10⁻⁷ - 1.78 × 10⁻⁷) M = (2.04 × 10⁻⁷ - 1.91 × 10⁻⁷) M = 0.13 × 10⁻⁷ M = 1.3 × 10⁻⁸ M

It can be expressed as : [H⁺] = 1.91 × 10⁻⁷ M ± 0.13 × 10⁻⁷ M
or : [H⁺] = 1.91 × 10⁻⁷ M ± 1.3 × 10⁻⁸ M

the answer is
.. 1.9x10^-7 M ± 0.1x10^-7 M

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there a couple of different ways you could approach this, I'll show you the easiest. You could also try formal propagation of errors.. (I outlined the general equation involving derivatives here https://answers.yahoo.com/question/index?qid=20160123100112AAZ1PoG )

first,... it's important to note that we can't use sig figs because sig figs assumes ± 1 in the right most significant digit. meaning had our value been 6.72 ± 0.01, we could use sig figs! but being that it's ± 0.03, we can't.

second, ... we know that pH = -log[H+] so that [H+] = 10^-pH

*** evaluating the max and min values ***

we know that 6.72 - 0.03 = 6.69
we know that 6.72 + 0.03 = 6.75

so we can plug those two values into [H+] = 10^-pH and calculate the results, determine the mean then +/- from those results
.. [H+] high = 10^-6.69 = 2.0 x 10^-7 M
.. [H+] low = 10^+6.75 = 1.8 x 10^-7 M

so the mean +/- would be 1.9x10^-7 M ± 0.1x10^-7 M

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note regarding the number of digits you should report to.

when evaluating log functions, the number of digits of the mantissa of the result should be the same as the precision of the number undergoing the log function.

example
.. log(3.254) = 0.5124... the number to the right of the decimal point in 0.5124 is the mantissa
... .. ..↑↑↑↑ .... .. .↑↑↑↑
.... . .4 digits... . 4 digits in the mantissa

our math here is the reverse
.. log(___.___) = 6.72
... . . .... .... .. ... .. . ↑↑
... ... ... .. .... .. .. .2 digits in the mantissa.. 72
so the results should be
.. [H+] = 10^(-6.99) = 2.0 x 10^-7 M
... . . .... .... .. ... .. ... ...↑↑
... ... ... .. .... .. .. .2 digits
and
.. [H+] = 10^(-6.99) = 1.8 x 10^-7 M
... . . .... .... .. ... .. ... ...↑↑
... ... ... .. .... .. .. .2 digits

so the final answer should be
.. 1.9x10^-7 M +/- 0.1x10^-7M
instead of
.. 1.91x10^-7 M +/- 0.13x10^-7 M

in addition, some instructors will tell you to limit the error to 1 digit.. 0.1 not 0.13. I usually accept either but almost always add the "some instructors might object" comment.