Answer
$$\rho = \frac{{3\sqrt 2 }}{2}$$
Work Step by Step
$$\eqalign{
& x = {e^t}\cos t,\,\,\,\,y = {e^t}\sin t,\,\,\,\,z = {e^t};\,\,\,\,\,\,\,t = 0 \cr
& {\text{Let }}{\bf{r}}\left( t \right) = x\left( t \right){\bf{i}} + y\left( t \right){\bf{j}} + z\left( t \right){\bf{k}} \cr
& {\text{Then}} \cr
& {\bf{r}}\left( t \right) = {e^t}\cos t{\bf{i}} + {e^t}\sin t{\bf{j}} + {e^t}{\bf{k}} \cr
& {\text{Calculate the derivatives }}{\bf{r}}'\left( t \right){\text{ and }}{\bf{r}}{\text{''}}\left( t \right){\text{ and evaluate at }}t = 0 \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {{e^t}\cos t} \right]{\bf{i}} + \frac{d}{{dt}}\left[ {{e^t}\sin t} \right]{\bf{j}} + \frac{d}{{dt}}\left[ {{e^t}} \right]{\bf{k}} \cr
& {\bf{r}}'\left( t \right) = \left( {{e^t}\cos t - {e^t}\sin t} \right){\bf{i}} + \left( {{e^t}\cos t + {e^t}\sin t} \right){\bf{j}} + {e^t}{\bf{k}} \cr
& {\bf{r}}'\left( t \right) = {e^t}\left( {\cos t - \sin t} \right){\bf{i}} + {e^t}\left( {\cos t + \sin t} \right){\bf{j}} + {e^t}{\bf{k}} \cr
& {\bf{r}}'\left( 0 \right) = {e^0}\left( {\cos 0 - \sin 0} \right){\bf{i}} + {e^0}\left( {\cos 0 + \sin 0} \right){\bf{j}} + {e^0}{\bf{k}} \cr
& {\bf{r}}'\left( 0 \right) = {\bf{i}} + {\bf{j}} + {\bf{k}} \cr
& and \cr
& {\bf{r}}''\left( t \right) = \frac{d}{{dt}}\left[ {{e^t}\left( {\cos t - \sin t} \right)} \right]{\bf{i}} + \frac{d}{{dt}}\left[ {{e^t}\left( {\cos t + \sin t} \right)} \right]{\bf{j}} + \frac{d}{{dt}}\left[ {{e^t}} \right]{\bf{k}} \cr
& {\bf{r}}''\left( t \right) = \left[ {{e^t}\left( { - \sin t - \cos t} \right) + {e^t}\left( {\cos t - \sin t} \right)} \right]{\bf{i}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\, + \left[ {{e^t}\left( {\cos t + \sin t} \right) + {e^t}\left( { - \sin t + \cos t} \right)} \right]{\bf{j}} + {e^t}{\bf{k}} \cr
& {\bf{r}}''\left( t \right) = \left( {{e^t}\cos t - {e^t}\sin t - {e^t}\sin t - {e^t}\cos t} \right){\bf{i}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\, + \left( {{e^t}\cos t + {e^t}\sin t + {e^t}\cos t - {e^t}\sin t} \right){\bf{j}} + {e^t}{\bf{k}} \cr
& {\bf{r}}''\left( t \right) = - 2{e^t}\sin t{\bf{i}} + 2{e^t}\cos t{\bf{j}} + {e^t}{\bf{k}} \cr
& {\bf{r}}''\left( 0 \right) = - 2{e^0}\sin 0{\bf{i}} + 2{e^0}\cos 0{\bf{j}} + {e^0}{\bf{k}} \cr
& {\bf{r}}''\left( 0 \right) = 2{\bf{j}} + {\bf{k}} \cr} $$
\[\begin{gathered}
{\text{Calculate the cross product }}{\mathbf{r}}'\left( 0 \right) \times {\mathbf{r}}''\left( 0 \right) \hfill \\
{\mathbf{r}}'\left( 0 \right) \times {\mathbf{r}}''\left( 0 \right) = \left| {\begin{array}{*{20}{c}}
{\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\
1&1&1 \\
0&2&0
\end{array}} \right| \hfill \\
{\mathbf{r}}'\left( 0 \right) \times {\mathbf{r}}''\left( 0 \right) = \left| {\begin{array}{*{20}{c}}
1&1 \\
2&1
\end{array}} \right|{\mathbf{i}} - \left| {\begin{array}{*{20}{c}}
1&1 \\
0&1
\end{array}} \right|{\mathbf{j}} + \left| {\begin{array}{*{20}{c}}
1&1 \\
0&2
\end{array}} \right|{\mathbf{k}} \hfill \\
{\mathbf{r}}'\left( 0 \right) \times {\mathbf{r}}''\left( 0 \right) = \left( {1 - 2} \right){\mathbf{i}} - \left( {1 - 0} \right){\mathbf{j}} + \left( {2 - 0} \right){\mathbf{k}} \hfill \\
{\mathbf{r}}'\left( 0 \right) \times {\mathbf{r}}''\left( 0 \right) = - {\mathbf{i}} - {\mathbf{j}} + 2{\mathbf{k}} \hfill \\
\end{gathered} \]
$$\eqalign{
& {\text{Use the formula }}\left( 3 \right)\,\,\,\kappa \left( t \right) = \frac{{\left\| {{\bf{r}}'\left( t \right) \times {\bf{r}}''\left( t \right)} \right\|}}{{{{\left\| {{\bf{r}}'\left( t \right)} \right\|}^3}}}.{\text{ Thus}}{\text{, at }}t = 0 \cr
& \kappa \left( 0 \right) = \frac{{\left\| { - {\bf{i}} - {\bf{j}} + 2{\bf{k}}} \right\|}}{{{{\left\| {{\bf{i}} + {\bf{j}} + {\bf{k}}} \right\|}^3}}} \cr
& \kappa \left( 0 \right) = \frac{{\sqrt {1 + 1 + 4} }}{{{{\left( {\sqrt {1 + 1 + 1} } \right)}^3}}} \cr
& \kappa \left( 0 \right) = \frac{{\sqrt 6 }}{{{{\left( {\sqrt 3 } \right)}^3}}} \cr
& \kappa \left( 0 \right) = \frac{{\sqrt 2 \sqrt 3 }}{{{{\left( 3 \right)}^{3/2}}}} \cr
& \kappa \left( 0 \right) = \frac{{\sqrt 2 }}{3} \cr
& \cr
& {\text{Calculate the radius of curvature }}\rho {\text{ at }}t = 0 \cr
& \rho = \frac{1}{{\kappa \left( 0 \right)}} = \frac{1}{{\sqrt 2 /3}} \cr
& \rho = \frac{3}{{\sqrt 2 }}\left( {\frac{{\sqrt 2 }}{{\sqrt 2 }}} \right) \cr
& \rho = \frac{{3\sqrt 2 }}{2} \cr} $$
