From Rutherford-Soddy law, the number of atoms left after $n$ half-lives is given by

$N=N_0{\left(\frac{1}{2}\right)}^n$

Where,  $N_0$ is the original number of atoms.

The number of half-life

$n=\frac{\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{e}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{d}\mathrm{e}\mathrm{c}\mathrm{a}\mathrm{y}}{\mathrm{e}\mathrm{f}\mathrm{f}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{ }\mathrm{h}\mathrm{a}\mathrm{l}\mathrm{f}-\mathrm{l}\mathrm{i}\mathrm{f}\mathrm{e}}$

Relation between effective disintegration constant ($\lambda$) and half-life ($T)$ is
$\lambda =\frac{\ln 2}{T}$

$\therefore {\lambda }_1+{\lambda }_2=\frac{\ln 2}{T_1}+\frac{\ln 2}{T_2}$

Effective half -life

$\frac{1}{T}=\frac{1}{T_1}+\frac{1}{T_2}=\frac{1}{1620}+\frac{1}{810}$

$\frac{1}{T}=\frac{1+2}{1620}\Rightarrow T\Rightarrow 540\mathrm{ }\mathrm{y}\mathrm{r}$

$\therefore n= \frac{t}{540}$

$\Rightarrow$ $\frac{t}{540}=2$

$\Rightarrow t= 2 \times 540 = 1080 \mathrm{y}\mathrm{r}$