I obtained the formula by differentiating it as follows where $s$ denotes the arc length parameter.
$T=\frac{dr}{ds}=\frac{dr}{dt}\frac{dt}{ds} \Leftrightarrow r'=s'T \\ r'' = s''T +s'T'\\\frac{dT}{dt}=\frac{dT}{ds}\frac{ds}{dt}=s'\kappa N \\ \therefore r''= s''T+(s')^2\kappa N \\ r'''=s'''T+s''T'+2s's''\kappa N+(s')^2\kappa 'N+(s')^2\kappa \frac{dN}{dt} \\ \frac{dN}{dt}=\frac{dN}{ds}\frac{ds}{dt} =(-\kappa T+\tau B)s'\\r'''=(s'''-(s')^3\kappa ^2)T+(3s’s''\kappa+(s')^2\kappa')N+(s')^3\kappa\tau B $
However, the coefficient of $N$ of $r'''$ is slightly different from deriving the formula of the torsion of a curve 's second answer (answered by Chappers).
It's too late to comment on that thread, so please forgive me for posting a question along with my solution. Is there anything wrong with my solution?
Barring typos in your attempt (such as missing $T$ in the first term and $s$ instead of $s'$ in $3ss''\kappa$), the only step that's needed is $\kappa'=\dot\kappa s'$ from the chain rule, which matches the original answer.