Answer
$$K = \frac{9}{{10\sqrt {10} }},{\text{ }}r = \frac{{10\sqrt {10} }}{9}$$
Work Step by Step
$$\eqalign{
& y = {e^{3x}},{\text{ }}x = 0 \cr
& {\text{Calculate the curvature, use }}K = \frac{{\left| {y''\left( x \right)} \right|}}{{{{\left( {1 + {{\left[ {y'\left( x \right)} \right]}^2}} \right)}^{3/2}}}} \cr
& {\text{Find }}y'\left( x \right){\text{ and }}y''\left( x \right),{\text{ }} \cr
& y'\left( x \right) = \frac{d}{{dx}}\left[ {{e^{3x}}} \right] \cr
& y'\left( x \right) = 3{e^{3x}} \cr
& {\text{Evaluate at }}x = 0 \cr
& y'\left( 0 \right) = 3 \cr
& y''\left( x \right) = \frac{d}{{dx}}\left[ {3{e^{3x}}} \right] \cr
& y''\left( x \right) = 9{e^{3x}} \cr
& {\text{Evaluate at }}x = 0 \cr
& y''\left( 0 \right) = 9 \cr
& \underbrace {K = \frac{{\left| {y''\left( x \right)} \right|}}{{{{\left( {1 + {{\left[ {y'\left( x \right)} \right]}^2}} \right)}^{3/2}}}}}_ \Downarrow \cr
& {\text{ at }}x = 0 \cr
& K = \frac{{\left| 9 \right|}}{{{{\left( {1 + {{\left[ 3 \right]}^2}} \right)}^{3/2}}}} \cr
& K = \frac{9}{{10\sqrt {10} }} \cr
& {\text{The radius of curvature is }}r = \frac{1}{K} \cr
& r = \frac{{10\sqrt {10} }}{9} \cr} $$
