Answer
${\bf 0.29}\;\rm T $
Work Step by Step
We have a coil that consists of $N$-loops (or turns) and we have here a constant flux from the magnet.
When the coil rotates, the angle between the flux and the area vector changes, so we must have an induced current and an induced emf.
We know that the induced current is given by
$$I_{\rm loop}=\dfrac{\varepsilon_{\rm loop }}{R_{\rm loop}}\tag 1$$
We know that the induced emf $\varepsilon$ is given by
$$\varepsilon_{\rm loop}=\left|\dfrac{d\Phi_{\rm }}{dt}\right|\tag 2$$
where $ \Phi =\vec A\cdot \vec B =AB\cos(2\pi f t) $
$$ \Phi = \pi r^2 B\cos(2\pi f t)$$
Plug into (2),
$$\varepsilon_{\rm loop}= \left|\dfrac{d }{dt} \pi r^2 B\cos(2\pi f t)\right| $$
And for $N$-turns,
$$\varepsilon_{\rm coil}= \pi Nr^2 B \left|\dfrac{d }{dt}\cos(2\pi f t)\right| $$
$$\varepsilon_{\rm coil}= (2\pi f) \pi N r^2 B \left|-\sin(2\pi f t)\right| $$
$$\varepsilon_{\rm coil}=2 \pi^2 r^2 f N B \sin(2\pi f t) $$
Amplitudes occurs at $\sin(2\pi f t) =1$, so
$$\varepsilon_{\rm max}=(2 \pi^2 r^2 Nf )B $$
The voltage measured by a standard voltmeter is the rms which is given by
$$\varepsilon_{\rm rms}=\dfrac{\varepsilon_{\rm max}}{\sqrt{2}}=\dfrac{2 \pi^2 r^2 Nf }{\sqrt{2}} B$$
$$\varepsilon_{\rm rms} =\dfrac{2 \pi^2 r^2 Nf }{\sqrt{2}} B$$
We can see that the emf is a function of $f$.
Plug the known;
$$\varepsilon_{\rm rms}=\dfrac{2 \pi^2 (1\times 10^{-2})^2 (100) B }{\sqrt{2}} f $$
$$\varepsilon_{\rm rms}=(0.14 B)f $$
Where the slope $=0.14 B$, Hence,
$$B=\dfrac{\rm Slope}{0.14}\tag 1$$
Now we need to plug the given data as dots where $y=\varepsilon_{\rm rms}$, and $x=f$, as shown below.
Plug the slope into (1),
$$B=\dfrac{0.04}{0.14}$$
$$B=\color{red}{\bf 0.29}\;\rm T $$
