Answer
${\bf+4.69}\;\rm T/s$
Work Step by Step
We have here a loop and a changing flux, so we must have an induced current.
We know that the induced current is given by
$$I=\dfrac{\varepsilon}{R}\tag 1$$
where the induced emf $\varepsilon$ is given by
$$\varepsilon=\left|\dfrac{d\Phi}{dt}\right|$$
where $\phi =AB\cos\theta$ and here $A$ is constant and $\theta= 0^\circ$, so that
$$\varepsilon=A\left|\dfrac{dB}{dt}\right|$$
where here $A=L^2$
$$\varepsilon=L^2\left|\dfrac{ dB}{dt}\right| $$
Hence, the rate of the magnetic field strength changing is given by
$$\dfrac{dB}{dt}=\dfrac{\varepsilon}{L^2}$$
Plug $\varepsilon$ from (1);
$$\dfrac{dB}{dt}=\dfrac{IR}{L^2}$$
Plug the known, from the given graph,
$$\dfrac{dB}{dt}=\dfrac{(150\times 10^{-3})(0.20)}{(0.08)^2}$$
$$\dfrac{dB}{dt}=\color{red}{\bf4.69}\;\rm T/s$$
As we see from the given figure, the induced current creates a magnetic field with a direction out of the page while the external magnetic field is into the page. We know that the induced flux is opposing the original flux, so the magnetic field increases.
