Step 1: Find how much reactant is left.
The reaction is $93.75\%$ complete, so the fraction remaining is \[ 100\% - 93.75\% = 6.25\%. \]
Step 2: Turn this into a simple fraction.
$6.25\% = \dfrac{6.25}{100} = \dfrac{1}{16}$, so one sixteenth of the reactant is left.
Step 3: Recall the half life behaviour of a first order reaction.
After each half life the amount of reactant is halved. So the fractions go $\dfrac{1}{2}, \dfrac{1}{4}, \dfrac{1}{8}, \dfrac{1}{16}$ after one, two, three and four half lives.
Step 4: Count the number of half lives.
Since $\dfrac{1}{16} = \left(\dfrac{1}{2}\right)^4$, exactly four half lives have passed to reach this point.
Step 5: Relate total time to the half life.
The total time for these four half lives is given as $x$ minutes, so \[ x = 4 \times t_{1/2}. \]
Step 6: Solve for the half life.
Rearranging, $t_{1/2} = \dfrac{x}{4}$. So the half life is
\[ \boxed{\dfrac{x}{4}} \]