Answer
$13$ years.
Work Step by Step
Compounding interest $n$ times a year is modeled by
$A=P(1+r/n)^{nt}$
Given $A=2P, r=0.055, n=2$, we find $t$
$2P=P(1+0.055/2)^{2t}$
$2=(1+0.055/2)^{2t}$
$\log 2=2t\cdot\log(1+0.055/2)$
$ t=\displaystyle \frac{\log 2}{2\log(1+0.055/2)}\approx~12.77517$
or, after about $13$ years.
