Answer
$43$
Work Step by Step
Using relation $T(t)=T_s+D_0e^{-kt}$ ...(1)
As we are given that $T(t)=90; T_s=60; k=0.0341$
and $D_0=190-60=130$
Equation (1), becomes
$90=60+130e^{-0.0341t}$
This can be solved as:
$e^{-0.0341t}=\dfrac{30}{130} \\ \implies -0.0341t=ln (\dfrac{30}{130})$
or, $t=\dfrac{1.466}{0.0341} \approx 43$
