Answer
\[
\begin{aligned}
\tau(t) &=\frac{a[x \sin t \hat{i}-c \cos t \hat{j}+a \hat{k}] \cdot(a \sin t \hat{i}-a \cos t \hat{j})}{a^{2}\left(a^{2}+c^{2}\right)} \\
&=\frac{c \cos ^{2} t+c \sin ^{2} t}{a^{2}+c^{2}}=\frac{c\left(\cos ^{2} t+\sin ^{2} t\right)}{c^{2}+a^{2}}=\frac{c}{c
^{2}+a^{2}}
\end{aligned}
\]
Work Step by Step
Given the curve $\mathbf{r}(t)=a \cos t \hat{i}+a \sin t \hat{j}+c t \hat{k},$
\[
\frac{\left[\mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime \prime}(t)\right] \cdot \mathbf{r}^{\prime \prime \prime}(t)}{\left\|\mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime \prime}(t)\right\|^{2}}=\tau(t)
\]
where
\[
\begin{aligned}
\mathbf{r}^{\prime}(t) &=\frac{d}{d t} \mathbf{r}(t)=-a \sin t \hat{i}+a \cos t j+c \hat{k} \\
\mathbf{r}^{\prime \prime}(t) &=\frac{d}{d t} \mathbf{r}^{\prime}(t)=-a \cos t \hat{i}-a \sin t \hat{j} \\
\mathbf{r}^{\prime \prime \prime}(t) &=\frac{d}{d t} \mathbf{r}^{\prime \prime}(t)=a \sin t \hat{i}-a \cos t \hat{j} \\
\mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime \prime}(t) &=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
-a \sin t & a \cos t & c \\
-a \cos t & -a \sin t & 0
\end{array}\right|=a[c \sin t \hat{i}-c \cos t \hat{j}+a \hat{k}(5)\\
\left\|\mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime \prime}(t)\right\|=a^{2}\left[c^{2}\left(\cos ^{2} t+\sin ^{2} t\right)+a^{2}\right]=\left(a^{2}+c^{2}\right) a^{2}
\end{aligned}
\]
Using all of these results, we find:
\[
\begin{aligned}
\tau(t) &=\frac{a[x \sin t \hat{i}-c \cos t \hat{j}+a \hat{k}] \cdot(a \sin t \hat{i}-a \cos t \hat{j})}{a^{2}\left(a^{2}+c^{2}\right)} \\
&=\frac{c \cos ^{2} t+c \sin ^{2} t}{a^{2}+c^{2}}=\frac{c\left(\cos ^{2} t+\sin ^{2} t\right)}{c^{2}+a^{2}}=\frac{c}{c
^{2}+a^{2}}
\end{aligned}
\]
