Answer
b. $I$ is from $a$ to $b$ and increasing.
f. $I$ is from $b$ to $a$ and decreasing.
Work Step by Step
There are two possible current answers; since the question doesn't specify which direction the current could be in, then there are two scenarios with the current running in each direction that we must consider.
Let us first consider $\displaystyle \frac{dI}{dt}$ when the current travels from $a$ to $b$. Since $a$ has a higher electric potential than $b$, there is a negative voltage drop across the inductor in this case. Thus, the inductor, which acts as a virtual emf, is creating an induced current that OPPOSES the current flow from $a$ to $b$ so the current must be increasing so cause there to be a negative induced current. Statement b. is true.
Now, let us consider what happens when the current flows from $b$ to $a$. Now, we go from the right side $b$, which has a lower electric potential than $a$, so we have a potential difference gain $\Delta V > 0$. This means that the induced current must be IN THE SAME DIRECTION as the regular current which means that the regular current $I$ must be decreasing. Statement f. is true.
b. $I$ is from $a$ to $b$ and increasing.
f. $I$ is from $b$ to $a$ and decreasing.
