Answer
The curvature at ${t_0}$ is
$\kappa \left( {{t_0}} \right) = 4.5$
Work Step by Step
At time ${t_0}$, we have the velocity vector ${\bf{v}} = 2{\bf{i}}$ and acceleration vector ${\bf{a}} = 3{\bf{i}} + 18{\bf{k}}$.
The unit tangent vector is
${\bf{T}} = \frac{{\bf{v}}}{{||{\bf{v}}||}} = \frac{{\left( {2,0,0} \right)}}{{\sqrt {\left( {2,0,0} \right)\cdot\left( {2,0,0} \right)} }} = \left( {1,0,0} \right)$
By Eq. (2) of Theorem 1 we have
${a_{\bf{T}}} = {\bf{a}}\cdot{\bf{T}} = \left( {3,0,18} \right)\cdot\left( {1,0,0} \right)$
${a_{\bf{T}}} = 3$
Next, we use Eq. (3) of Theorem 1 to find
${a_{\bf{N}}}{\bf{N}} = {\bf{a}} - {a_{\bf{T}}}{\bf{T}}$
${a_{\bf{N}}}{\bf{N}} = \left( {3,0,18} \right) - 3\left( {1,0,0} \right)$
${a_{\bf{N}}}{\bf{N}} = \left( {0,0,18} \right)$
Since ${\bf{N}}$ is an unit vector, so
${a_{\bf{N}}} = ||{a_{\bf{N}}}{\bf{N}}|| = 18$
At time ${t_0}$, the velocity vector is ${\bf{v}} = 2{\bf{i}}$, so the speed is $v=2$.
By Eq. (1):
${a_{\bf{N}}} = \kappa \left( t \right)v{\left( t \right)^2}$
the curvature at ${t_0}$ is
$\kappa \left( {{t_0}} \right) = \frac{{18}}{{{2^2}}} = 4.5$
