Explanation:
The radius of curvature is
\(\rho = \frac{{{{\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]}^{3/2}}}}{{\left| {\frac{{{d^2}y}}{{d{x^2}}}} \right|}}\)
\(y = 3{x^2}\)
\(\frac{{dy}}{{dx}} = 6x,\;\,\frac{{{d^2}y}}{{d{x^2}}} = 6\)
\(x = y = 0 \Rightarrow \frac{{dy}}{{dx}} = 0\,\,\& \,\,\frac{{{d^2}y}}{{d{x^2}}} = 6\)
\(\rho = 1/6\)