Answer
See the detailed answer below.
Work Step by Step
Let's assume that the current moves from the left to the right through the inductor.
We know that the induced potential difference is given by
$$\Delta V=-L\dfrac{dI}{dt}$$
Hence,
$$\Delta V=-L\dfrac{I_f-I_i}{\Delta t}$$
Plug the known;
$$\Delta V=-(10\times 10^{-3})\dfrac{(50-150)\times 10^{-3}}{10\times 10^{-6}}$$
$$\Delta V=\color{red}{\bf 100}\;\rm V$$
The decrease of the current decreases the magnetic flux of the inductor.
According to Lenz's law, the induced current will oppose this decrease of the magnetic flux inside the loop. This means that the induced current will be in the same direction as the original one.
Hence, the induced potential difference $\bf increases$ along the direction of the original current.
