Because 25% is half of 0.5 and the half life of [tex]{}^{14}\text{C}[/tex] is given as 5,715, the time it takes to reach this amount should just be double the half life: 11430 years.
However, in other cases the problem may not have easy numbers, so below is a more mathematical approach.
The half life formula is [tex]A=A_0*(\frac{1}{2} )^{\frac{t}{h} }[/tex].
Solving for t gives:
[tex]t=\frac{h*\ln(\frac{A}{A_0} )}{\ln(\frac{1}{2} )}[/tex]
Given information:
[tex]A=0.25A_0\\h=5715\text{ yrs}[/tex]
Plug in given values and solve for t.
[tex]t=\frac{5715 *\ln(0.25) }{\ln(\frac{1}{2} )} =11430 \text{ yrs}[/tex]
With correct significant figures, this would be 11000 years.
