When you say the point (4π,1), those appear to be (x,y) coordinates. From the x expression x=8t or t=x/8.

1) Therefore the t value that corresponds to x = 4π is t = π/2

2) The equation for curvature is |v x a|/ |v • v|^{3/2}

curvature is |(8,5cos(5t)) x (0,-25sin(5t))/ (64 +cos^{2}(5t))^{1.5}

8*(25)|sin(5t)|/(64+cos^{2}(5t))^{1.5}

Now insert t=pi/2 : 8*(25)/64^{1.5} or 25/64. The radius is 1/κ

Hope that helps. I checked the result with Wolfram Alpha.