If the compound interest on $1000$ art the rate of $10\%p.a$ be $210$ then the

time is 

$P=1000,R=10\%,C.I=210$

$C.I=A-P=P{(1+\frac{R}{100})}^n-P$

$=P[{(1+\frac{R}{100})}^n-1]$

$210=1000[{(1+\frac{10}{100})}^n-1]$

$\frac{21}{100}=[{\left(\frac{11}{10}\right)}^n-1]$

$\Rightarrow$$\frac{21}{100}+1={\left(\frac{11}{10}\right)}^n$

$\frac{121}{100}={\left(\frac{11}{10}\right)}^n\Rightarrow {\left(\frac{11}{10}\right)}^2={\left(\frac{11}{10}\right)}^n$

$n=2$

$\therefore$ Time $=2years$