Answer
\[
\begin{array}{l}
\tau(t)=\frac{\left[\mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime \prime}(t)\right] \cdot \mathbf{r}^{\prime \prime \prime}(t)}{\left\|\mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime \prime}(t)\right\|}=\frac{\left(2 t^{2} \mathbf{i}-4 t \mathbf{j}+4 \mathbf{k}\right) \cdot(2 \mathbf{k})}{2\left(2+t^{2}\right)^{2}} \\
\left[\frac{2}{(2+t^{2})^2}\right]=\tau(t)
\end{array}
\]
Work Step by Step
Given the curve $2 t \mathbf{i}+t^{2} \mathbf{j}+\frac{1}{3} t^{3} \mathbf{k}=\mathbf{r}(t)$
\[
\begin{aligned}
2 \mathbf{i}+2 t \mathbf{j}+t^{2} \mathbf{k} &=\frac{d}{d t} \mathbf{r}(t)=\mathbf{r}^{\prime}(t) \\
&\frac{d^{2}}{d t^{2}} \mathbf{r}(t)=\frac{d}{d t} \mathbf{r}^{\prime}(t)=2 \mathbf{j}+2 t \mathbf{k}=\mathbf{r}^{\prime \prime}(t)
\end{aligned}
\]
Using (1) and (2) we get that:
\[
\mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime \prime}(t)=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & 2 t & t^{2} \\
0 & 2 & 2 t
\end{array}\right|=\left(4 t^{2}-2 t^{2}\right) \mathbf{i}-4 t \mathbf{j}+4 \mathbf{k}=2 t^{2} \mathbf{i}-4 t \mathbf{j}+4 \mathbf{k}
\]
where
\[
\left\|\mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime \prime}(t)\right\|=(\sqrt{16+4 t^{4}+16 t^{2}})^{2}=2\left(2+t^{2}\right)^{2}
\]
Furthemore
\[
\frac{d}{d t} \mathbf{r}^{\prime \prime}(t)=2 k=
\mathbf{r}^{\prime \prime \prime}(t)
\]
Using the formula in $d$ :
\[
\begin{array}{l}
\tau(t)=\frac{\left[\mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime \prime}(t)\right] \cdot \mathbf{r}^{\prime \prime \prime}(t)}{\left\|\mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime \prime}(t)\right\|}=\frac{\left(2 t^{2} \mathbf{i}-4 t \mathbf{j}+4 \mathbf{k}\right) \cdot(2 \mathbf{k})}{2\left(2+t^{2}\right)^{2}} \\
\left[\frac{2}{(2+t^{2})^2}\right]=\tau(t)
\end{array}
\]
