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