Answer
See the detailed answer below.
Work Step by Step
We know from, Faraday’s law of electric induction, that
$$\oint \vec E\cdot d\vec s=A\left| \dfrac{dB}{dt}\right|$$
As we see below, the electric field is tangent to the curve at $r\gt R$ at any point on its surface.
Hence,
$$ E (2\pi r)=A\left| \dfrac{dB}{dt}\right|$$
$$ E (2\pi r)=\pi R^2\left| \dfrac{dB}{dt}\right|$$
Thus,
$$\boxed{E=\dfrac{R^2}{2r}\left| \dfrac{dB}{dt}\right|}$$
To be sure, let's test at $r=R$,
$$E=\dfrac{R }{2}\left| \dfrac{dB}{dt}\right|$$
which is the strength of the induced electric field inside the solenoid.