Answer
a) ${\bf 0.94}\;\rm V/m$
b) ${\bf 10}\;\rm T$
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the strength of the induced electric field inside the solenoid is given by
$$E=\dfrac{r}{2}\left| \dfrac{dB}{dt}\right|\tag 1$$
where $B$ here varies sinusoidally between 8.0 T and 12.0 T at a frequency of 10 Hz, so
$$B=10+2\sin(2\pi f t)\tag 2$$
Plug into (1),
$$E=\dfrac{r}{2}\left| \dfrac{d }{dt}\left[10+2\sin(2\pi f t)\right]\right| $$
$$E=\dfrac{r}{2}\left| 4\pi f\cos(2\pi f t) \right| $$
$$E=\dfrac{4\pi f r}{2}\left| \cos(2\pi f t) \right| $$
$$E=2\pi f r\left| \cos(2\pi f t) \right| $$
So the maximum electric field occurs at $\cos(2\pi ft)=\pm 1$, so
$$E_{\rm max}=2\pi f r $$
Plug the known;
$$E_{\rm max}=2\pi (10)(1.5\times 10^{-2})$$
$$E_{\rm max}=\color{red}{\bf 0.94}\;\rm V/m$$
$$\color{blue}{\bf [b]}$$
We need $B$ at $E_{\rm max}$ where $\cos(2\pi f t)=1$ which means $2\pi ft =0$.
Plug that into (2),
$$B=10+2\sin(0) $$
$$B=\color{red}{\bf 10}\;\rm T$$
