Answer
$$K = \frac{2}{{5\sqrt 5 }},{\text{ }}r = \frac{{5\sqrt 5 }}{2}$$
Work Step by Step
$$\eqalign{
& y = {e^{ - x/2}},{\text{ }}x = 0 \cr
& {\text{Calculate the curvature, use }}K = \frac{{\left| {y''\left( x \right)} \right|}}{{{{\left( {1 + {{\left[ {y'\left( x \right)} \right]}^2}} \right)}^{3/2}}}} \cr
& {\text{Find }}y'\left( x \right){\text{ and }}y''\left( x \right){\text{}} \cr
& y'\left( x \right) = \frac{d}{{dx}}\left[ {{e^{ - x/2}}} \right] \cr
& y'\left( x \right) = - \frac{1}{2}{e^{ - x/2}} \cr
& {\text{Evaluate at }}x = 0 \cr
& y'\left( 0 \right) = - \frac{1}{2}{e^0} = - \frac{1}{2} \cr
& y''\left( x \right) = \frac{d}{{dx}}\left[ { - \frac{1}{2}{e^{ - x/2}}} \right] \cr
& y''\left( x \right) = \frac{1}{4}{e^{ - x/2}} \cr
& {\text{Evaluate at }}x = 0 \cr
& y''\left( 0 \right) = \frac{1}{4}{e^0} = \frac{1}{4} \cr
& \underbrace {K = \frac{{\left| {y''\left( x \right)} \right|}}{{{{\left( {1 + {{\left[ {y'\left( x \right)} \right]}^2}} \right)}^{3/2}}}}}_ \Downarrow \cr
& {\text{ at }}x = 0 \cr
& K = \frac{{\left| {\frac{1}{4}} \right|}}{{{{\left( {1 + {{\left[ {1/2} \right]}^2}} \right)}^{3/2}}}} \cr
& K = \frac{{1/4}}{{{{\left( {5/4} \right)}^{3/2}}}} \cr
& K = \frac{{\frac{1}{4}}}{{\frac{{5\sqrt 5 }}{8}}} \cr
& K = \frac{2}{{5\sqrt 5 }} \cr
& {\text{The radius of curvature is }}r = \frac{1}{K} \cr
& r = \frac{{5\sqrt 5 }}{2} \cr} $$
