Answer
$E = 7.15\times 10^{-5}~V/m$
Work Step by Step
Note that $r = 2.20~cm$ is less than the radius of the solenoid.
We can find the magnitude of the induced electric field:
$2\pi~r~E = \vert \frac{d\Phi}{dt} \vert$
$2\pi~r~E = (6.50~mT/s)~(\pi~r^2)$
$E = \frac{(6.50~mT/s)~(r)}{2}$
$E = \frac{(6.50~mT/s)~(0.022~m)}{2}$
$E = 7.15\times 10^{-5}~V/m$
