Solution

The magnetic force $evB$ provides centripetal acceleration to the electron, so

$$evB=\frac{mv^2}{r}\qquad\Rightarrow\qquad v=\frac{eBr}{m},$$

where $-e$, $m$, and $v$ are the charge, mass, and speed of the electron around the circle, respectively. Since $B=\alpha t$, the tangential acceleration of the electron

$$a=\frac{dv}{dt}=\frac{d}{dt}\left(\frac{eBr}{m}\right)=\frac{e\alpha r}{m}$$

It is caused by induced electric field $E$ as a result of changing magnetic flux. Thus,

$$a=\frac{eE}{m}\qquad\Rightarrow\qquad E=\alpha r.$$

Let $B'$ be the magnetic field inside the cylinder in the same direction as $B$. By the law of induction,

$$E\cdot2\pi r=\frac{d}{dt}(B'\cdot\pi r_0^2+B\cdot\pi(r^2-r_0^2)).$$

As a result,

$$\frac{dB'}{dt}=\alpha\left[\left(\frac{r}{r_0}\right)^2+1\right].$$

Since $B'$ is initially zero, we have $B'=\alpha t[(r/r_0)^2+1]$.