Answer
$$K = \frac{3}{{16}},{\text{ }}r = \frac{{16}}{3}$$
Work Step by Step
$$\eqalign{
& y = \frac{3}{4}\sqrt {16 - {x^2}} ,{\text{ }}x = 0 \cr
& {\text{Calculate the curvature, use }}K = \frac{{\left| {y''\left( x \right)} \right|}}{{{{\left( {1 + {{\left[ {y'\left( x \right)} \right]}^2}} \right)}^{3/2}}}} \cr
& {\text{Find }}y'\left( x \right){\text{ and }}y''\left( x \right),{\text{ }} \cr
& y'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{3}{4}\sqrt {16 - {x^2}} } \right] \cr
& y'\left( x \right) = \frac{3}{4}\left( {\frac{{ - 2x}}{{2\sqrt {16 - {x^2}} }}} \right) \cr
& y'\left( x \right) = \frac{{ - 3x}}{{4\sqrt {16 - {x^2}} }} \cr
& {\text{Evaluate at }}x = 0 \cr
& y'\left( 0 \right) = 0 \cr
& \cr
& y''\left( x \right) = - \frac{3}{4}\frac{d}{{dx}}\left[ {\frac{x}{{\sqrt {16 - {x^2}} }}} \right] \cr
& y''\left( x \right) = - \frac{3}{4}\left( {\frac{{\sqrt {16 - {x^2}} - x\left( {\frac{{ - 2x}}{{2\sqrt {16 - {x^2}} }}} \right)}}{{16 - {x^2}}}} \right) \cr
& {\text{Evaluate at }}x = 0 \cr
& y''\left( 0 \right) = - \frac{3}{4}\left( {\frac{{\sqrt {16} }}{{16}}} \right) \cr
& y''\left( 0 \right) = - \frac{3}{{16}} \cr
& \underbrace {K = \frac{{\left| {y''\left( x \right)} \right|}}{{{{\left( {1 + {{\left[ {y'\left( x \right)} \right]}^2}} \right)}^{3/2}}}}}_ \Downarrow \cr
& {\text{ at }}x = 0 \cr
& K = \frac{{\left| { - \frac{3}{{16}}} \right|}}{{{{\left( {1 + {{\left[ 0 \right]}^2}} \right)}^{3/2}}}} \cr
& K = \frac{3}{{16}} \cr
& {\text{The radius of curvature is }}r = \frac{1}{K} \cr
& r = \frac{{16}}{3} \cr} $$
