Answer: \(\boxed{20\ \mathrm{m\ s^{-2}}}\).

The transverse component of acceleration is the rate of change of the speed along the circle.

The speed is

\[v=8-pt+t^2.\]

So

\[\frac{dv}{dt}=-p+2t.\]

At \(t=2\), the transverse component is zero, hence

\[-p+2(2)=0.\]

Therefore

\[p=4.\]

Now find the speed at \(t=2\):

\[v=8-4(2)+2^2=8-8+4=4.\]

The radial component of acceleration is

\[a_R=\frac{v^2}{r}.\]

Here \(r=0.8\), so

\[a_R=\frac{4^2}{0.8}=\frac{16}{0.8}=20.\]

Hence the magnitude of the radial component is

\[\boxed{20\ \mathrm{m\ s^{-2}}}.\]