Let $N_A$ and $N_B$ be the number of molecules of $A$ and $B$ after time $t$.

Also, after time $t$, $\frac{N_B}{N_A}=0.3$(given).

Also, let $N_0$ be the total number of nucleus initial

After time $t$, $N_A+N_B=N_0$

$N_A+0.3N_A=N_0$

$\therefore N_A=\frac{N_0}{1.3}$

Also, the rate of disintegration of $A$

$N_A=N_0{e}^{-\lambda t}$

$\Rightarrow \frac{N_0}{1.3}=N_0{e}^{-\lambda t};\frac{1}{1.3}=e^{-\lambda t}or\log \left(1.3\right)=\lambda t$

$t=\frac{log(1.3)}{\lambda }$

$\therefore t=\frac{Tlog(1.3)}{log\left(2\right)}=\frac{Tlog(1.3)}{log2}$.