#### Answer

$\dfrac{3}{2^{3/2}}$

#### Work Step by Step

The curvature $\kappa$ for a plane curve system is:
$\kappa (t)= \dfrac{||r'(t) \times r''(t)||}{||r'(t)||^3}$
We have:
$r'(t) =\lt -1, e^{t-4}, 8-2t \gt$ and $r''(t) =\lt 0, e^{t-4}, -2 \gt$
Thus,
$\kappa(t) = \dfrac{||(2te^{t-4}-10e^{t-4})i-2j-e^{t-4}k||}{||\lt -1, e^{t-4}, 8-2t \gt||^3} \\=\dfrac{\sqrt {(2te^{t-4}-10e^{t-4})^2+4+e^{2t-8}}}{[1+e^{2t-8}+(8-2t)^2]^{3/2}}$
Now, we will compute $\kappa (t)$ at $t=4$
Therefore,
$\kappa(t)= \dfrac{\sqrt {(-2)^2+4+1}}{(1+1)^{3/2}}=\dfrac{3}{2^{3/2}}$