The particle has an acceleration vector with one component directed toward the center of its orbit, and the other directing tangentially to its orbit. Call these components [tex]\vec a_c[/tex] ([tex]c[/tex] for center) and [tex]\vec a_t[/tex] ([tex]t[/tex] for tangent). Then its acceleration vector has magnitude
[tex]|\vec a|=\sqrt{\|\vec a_c\|^2+\|\vec a_t\|^2}[/tex]
We have
[tex]\|\vec a_c\|=\dfrac{\|\vec v\|^2}r[/tex]
where [tex]\|\vec v\|[/tex] is the particle's speed and [tex]r[/tex] is the radius of orbit, so
[tex]\|\vec a_c\|=\dfrac{\left(37.2\frac{\rm m}{\rm s}\right)^2}{21\,\rm m}=65.9\dfrac{\rm m}{\mathrm s^2}[/tex]
We're given that the particle's speed changes at a rate of 23.1 m/s^2. Its velocity vector points in the same direction as [tex]\vec a_t[/tex], i.e. perpendicular to [tex]\vec a_c[/tex], so
[tex]\|\vec a_t\|=23.1\dfrac{\rm m}{\mathrm s^2}[/tex]
Then the magnitude of the particle's acceleration is
[tex]\|\vec a\|=\sqrt{\left(65.9\dfrac{\rm m}{\mathrm s^2}\right)^2+\left(23.1\dfrac{\rm m}{\mathrm s^2}\right)^2}=\boxed{69.8\dfrac{\rm m}{\mathrm s^2}}[/tex]
