#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#