#t = 0.515s# (for A) and #t = 1.74s# (for B).
The general form of a first order reaction is
#aA -> PRODUCTS#, with the reaction rate being
#rate = k * [A]#.
The equation we'll use for this problem is
#ln(([A])/[A]_@) = -kt#, where
#[A]_0# - the initial concentration of #A#;
#[A]# - the final concentration of #A#;
#k# - the rate constant - in your case equal to #0.693s^-1#
#t# - the reaction time;
Now, #30%# decomposition means that the final concentration of #A# will equal #70%# of the initial concentration - 30% consumed, 70% remaining. This can be written as
#[A] = (100 - 30)/100 * [A]_0 = 7/10 * [A]_0#
The equation becomes
#ln((7/10 * [A]_0)/[A]_0) = -k * t#
#ln(7/10) = -0.693 * t -> t = ln(0.7)/(-0.693 s^-1) = 0.515 s#
The same approach can be used for when #70%# decomposition occurs:
#[A] = (100-70)/100 * [A]_0 = 3/10 * [A]_0#
#ln((3/10 * [A]_0)/[A]_0) = -kt -> t = ln(0.3)/(-0.693s^-1) = 1.74s#
