Curvature at a point in a vector valued function

Question

I am trying to determine the curvature when $t=2$ of the function $r(t)=<t^3,3t^2,8t>$

So I found $v(t)=<3t^2,6t,8>$ and $a(t)=<6t,6,0>$. So now that I have these two functions, I can just use the formula $\kappa=\frac{||v(t) \times a(t)||}{||v(t)||^3}$

I find that $v(2) \times a(2)=<-48,96,-72>$ and $||v(2)||=18.76$ So according to the formula, $\kappa = \frac{\sqrt{48^2+96^2+72^2}}{18.76^3}=0.02$. According to the answer it should be the inverse of what I got. Where did I go wrong?

Answer 1

Is the question is asking for radius of curvature? since that is the reciprocal of curvature.