Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
From the given, we can see that the sinusoidal magnetic field inside the solenoid is given by
$$B=10+2\sin(2\pi f t)$$
where $f=10$ Hz, so
$$B=10+2\sin(20\pi t)\tag1$$
According to the Faraday’s law,
$$\oint \vec E\cdot d\vec s=\dfrac{-d\Phi_m}{dt}$$
If we imagined a circle of radius $r$ from the axis of the solenoid where $r\lt R_{\rm solenoid}$, then the electric field vectors are tangent to the circle at any point according to the right-hand rule.
So, $\oint \vec E\cdot d\vec s=E(2\pi r)$, and hence,
$$E(2\pi r) =\dfrac{-d\Phi_m}{dt}$$
Recalling that $\Phi_m= BA\cos\theta=BA\cos0^\circ$, so
$$E(2\pi r) =\dfrac{-d(BA)}{dt}$$
where $A=\pi r^2$
$$E(2\color{red}{\bf\not}\pi \color{red}{\bf\not}r) =\color{red}{\bf\not}\pi r^{\color{red}{\bf\not}2}\dfrac{-dB}{dt}$$
Plug $B$ from (1),
$$E =\dfrac{- r}{2} \dfrac{ d }{dt}\left[10+2\sin(20\pi t)\right]$$
$$E =\dfrac{- r}{2} \left[0+2(20\pi)\cos(20\pi t)\right]$$
$$\boxed{E = -20\pi r \cos(20\pi t) }$$
The maximum electric field occurs when $\cos(20\pi t)=-1$
$$ E_{max} = 20\pi r =20\pi (1.5\times 10^{-2}) $$
$$ E_{max} = \color{red}{\bf 0.94}\;\rm V/m$$
$$\color{blue}{\bf [b]}$$
At $ E_{max}$, $\cos(20\pi t)=-1$, hence, $\sin(20\pi t)=0$
Plug that into (1),
$$B=10+2(0)$$
$$B=\color{red}{\bf10}\;\rm T$$
