Answer
$2.67\times10^{-3}$ Tesla per second
Work Step by Step
Faraday's law can be written as
$\varepsilon_{induced} = \displaystyle |\frac{d\Phi_B}{dt}| = |\vec{A}\frac{d\vec{B}}{dt} + \vec{B}\frac{d\vec{A}}{dt}| = \oint\vec{E}\cdot d\vec{s}$
where $\vec{s}$ is the arc length. In this problem, the cross-sectional area is not changing so $\displaystyle \frac{d\vec{A}}{dt} = 0$ which leaves us with
$\displaystyle \varepsilon = \oint\vec{E}\cdot d\vec{s} = \vec{A}\frac{d\vec{B}}{dt}$
Solving for $\displaystyle \frac{d\vec{B}}{dt}$ gives us (since $\vec{s}$ is arc length, the arc length of an entire circle is equal to its circumference, $2\pi r$):
$\displaystyle \vec{E}\vec{s}= \vec{A}\frac{d\vec{B}}{dt}$
$\displaystyle \frac{\vec{E}\vec{s}}{\vec{A}}= \frac{\vec{E}(2\pi r)}{\pi r^2} = \frac{2\vec{E}}{r} = \frac{d\vec{B}}{dt}$
$\displaystyle \frac{d\vec{B}}{dt} = \frac{2(2\times10^{-3})}{1.5}= 2.67\times10^{-3}$ Tesla per second
