At a given instant, say t = 0, two radioactive substance A and B...

Given,
Half life of $A = \ln 2$
Initial number of nucleus is, ${R}_{0A} = {R}_{0B} = {R}_{0}$

Half life is related with decay constant as, ${\left({t}_{1/2}\right)}_{A} = \dfrac{\ln 2}{\lambda}$
${\lambda}_{A} = 1$

At $t = 0{R}_{A} = {R}_{B}$
From activity relation, we have,
${R}_{0}{e}^{- {\lambda}_{A}T} = {R}_{0}{E}^{- {\lambda}_{B}T}$

$\text{at}T = t, ⇒ \dfrac{{R}_{B}}{{R}_{A}} = \dfrac{{R}_{0}{e}^{- {\lambda}_{B}t}}{{R}_{0}{e}^{- {\lambda}_{A}t}}$

As per given question, $\dfrac{{R}_{B}}{{R}_{A}} = {e}^{- 3t}$, On comparing with above relation, we have
$⇒ {e}^{- \left({\lambda}_{B} - {\lambda}_{A}\right)t} = {e}^{- t}$
$⇒ {\lambda}_{B} - {\lambda}_{A} = 3$
$⇒ {\lambda}_{B} = 3 + 1 = 4$
$⇒ {\left({t}_{1/2}\right)}_{B} = \dfrac{\lambdan2}{{\lambda}_{B}} = \dfrac{\lambdan2}{4}$