Answer
$E = 1.43\times 10^{-4}~V/m$
Work Step by Step
Let $r_1 = 6.00~cm$, the radius of the solenoid.
Let $r_2 = 8.20~cm$
We can find the magnitude of the induced electric field:
$2\pi~r_2~E = \vert \frac{d\Phi}{dt} \vert$
$2\pi~r_2~E = (6.50~mT/s)~(\pi~r_1^2)$
$E = \frac{(6.50~mT/s)~(r_1)^2}{2r_2}$
$E = \frac{(6.50~mT/s)~(0.060~m)^2}{(2)(0.0820~m)}$
$E = 1.43\times 10^{-4}~V/m$
