Solution

The relation between the circulation of the magnetic induction around the loop and the electric displacement flux through the surface "stretched" over this loop is given by the equation:
$$\oint \vec B \, d\vec l = \mu_0 \varepsilon_0 \frac{d}{dt} \left( \int \vec E \, d\vec S \right).$$
Or
$$C_B = \mu_0 \varepsilon_0 \frac{d\Phi_E}{dt},$$
where $C_B$ is the circulation of the magnetic induction, and $\Phi_E$ is the electric displacement flux.

According to the condition ($t_0 = 1\ \mu s$),
$$\Phi_E(t) = kt + b, \qquad \Phi_E(t_0) = 0 \ \Rightarrow\ \Phi_E(t) = kt - k t_0.$$
Then, taking the time derivative:
$$C_B = \mu_0 \varepsilon_0 k,$$
$$\Phi_E(t) = \frac{C_B}{\mu_0 \varepsilon_0} (t - t_0) = C_B c^2 (t - t_0).$$
$$\Phi_E(0) = 1 \cdot 10^{-3}\ (\text{T}\cdot\text{m}) \cdot (3 \cdot 10^8\ (\text{m/s}))^2 \cdot 1 \cdot 10^{-6}\ (\text{s}) = 9 \cdot 10^7\ (\text{V}\cdot\text{m}).$$
Savchenko's answer contains a misprint.