According to radioactive decay law

$N=N_{e} e^{-\lambda t}$

where $N_{0}=$ Number of radioactive nuclei at time $t=0$

$N=$ Number of radioactive nuclei left undecayed at any time $t$

$\lambda=$ decay constant

At time $t_{2}, \frac{2}{3}$ of the sample had decayed

$\therefore N=\frac{1}{3} N_{0}$

$\therefore \frac{1}{3} N_{0}=N_{0} e^{-\lambda t_{2}}$  …… $(i)$

At time $t_{1}, \frac{1}{3}$ of the sample had decayed,

$\therefore N=\frac{2}{3} N_{0}$

$\therefore \frac{2}{3} N_{0}=N_{0} e^{-\lambda t_{1}}$  …… $(ii)$

Divide $(i)$ by $(ii)$, we get

$\frac{1}{2} = \frac{{{e^{ – \lambda {t_2}}}}}{{{e^{ – \lambda {t_1}}}}}$ $ \Rightarrow \frac{1}{2} = {e^{ – \lambda \left( {{t_2} – {t_1}} \right)}}$

${\lambda\left(t_{2}-t_{1}\right)=\ln 2} $

${t_2} – {t_1} = \frac{{\ln 2}}{\lambda } = \frac{{\ln 2}}{{\left( {\frac{{\ln 2}}{{{T_{1/2}}}}} \right)}}$ $\left( {\because \lambda  = \frac{{\ln 2}}{{{T_{1/2}}}}} \right)$

${=T_{10}=50 \text { days }}$