To answer this question we should know following key points:

1. Total Acceleration is given as $a_{total}=\sqrt{\left(a_t\right)^2+\left(a_n\right)^2}$

2. $a_t=\dfrac{dv}{dt} \text{ and } a_n=\dfrac{v^2}{\rho}$

Here, $s=0.5t^3+3t \text{ m }$

$\therefore v=\dfrac{ds}{dt}=1.5t^2+3 \text{ m/s}$

and 

$a_t=\dfrac{dv}{dt}=3t \text{ m/$s^2$}$ Hence, at t = 2 sec, $v=1.52^2+2=9 \text{ m/s }$ and  $a_t=32=6 \text{ m/$s^2$}$ But, given that at t = 2 sec., total acceleration $a_{total}=10 \text{ m/$s^2$}$ $\therefore 10= \sqrt{ 6^2 + a_n^2 }\ \therefore a_n=8 \text{ m/$s^2$}$ Also,  $a_n=\dfrac{v^2}{\rho}\ \therefore 8=\dfrac{9^2}{\rho}\ \therefore \rho=10.125 m$

Answer : Radius of curvature =10.125 m.