Answer
$0.2$ years
Work Step by Step
The model for compounding n times a year is
$A=P(1+\displaystyle \frac{r}{n})^{nt}$
We are given
$\left\{\begin{array}{l}
A=1200,\\
P=1000,\\
r=1.00,\\
n=4
\end{array}\right.\qquad t=?$
We solve for $t$ as follows:
$1200=1000(1+\displaystyle \frac{1.00}{4})^{4t}$
$\displaystyle \frac{1200}{1000}=(1.25)^{4t}$
$1.2=(1.25)^{4t}$
$\ln 1.2=4t\cdot\ln(1.25)$
$t=\displaystyle \frac{\ln 1.2}{4\ln(1.25)}\approx 0.2046\approx 0.2$ years
