Regular Expression-$a^{*}(a+ba)^{*}b^{*}$
0 length string-$\epsilon{\color{Green} \checkmark}$
1 Length string-
$a\ \ \ accepted{\color{Green} \checkmark}$
$b\ \ \ accepted{\color{Green} \checkmark}$
2 length string-
$aa\ \ \ accepted{\color{Green} \checkmark}$
$ab\ \ \ accepted{\color{Green} \checkmark}$
$ba\ \ \ accepted{\color{Green} \checkmark}$
$bb\ \ \ accepted{\color{Green} \checkmark}$
3 length string-
$aaa\ \ \ accepted{\color{Green} \checkmark}$
$aab\ \ \ accepted{\color{Green} \checkmark}$
$aba\ \ \ accepted{\color{Green} \checkmark}$
$baa\ \ \ accepted{\color{Green} \checkmark}$
$bab\ \ \ accepted{\color{Green} \checkmark}$
$bba\ \ \ not \ accepted{\color{Red} \times}$
$abb\ \ \ accepted{\color{Green} \checkmark}$
$bbb\ \ \ accepted{\color{Green} \checkmark}$
Here we can see 3 length string $bba$ can't be accepted by given regular expression.
Hence, length of the shortest string not in the language over the alphabet${a,b}$ of the regular expression $a^{*}(a+ba)^{*}b^{*}$ is 3.