Voltage drop across an inductor 
An inductor should oppose the current right, so shouldn't the voltage drop across the inductor be in the opposite direction as that shown in the figure?
 A: 
An inductor should oppose the current right

No, that's not right.  An (ideal) inductor 'opposes' a change in current but offers no opposition to a fixed current, i.e., the voltage $e$ across the inductor is zero for constant $I$.

so shouldn't the voltage drop across the inductor be in the opposite
  direction as that shown in the figure?

The $+$ and $-$ signs are there to indicate the reference polarity of the voltage across the inductor.  The voltage $e$ can be positive or negative and, according to the drawing, if $e$ is positive, the left-most inductor terminal is positive with respect to the right-most terminal.  But, if $e$ is negative, the left-most terminal is negative with respect to the right-most terminal.
As the circuit is drawn, if $I$ is increasing, the inductor voltage $e$ should be positive to oppose the increase of $I$.

From the comments:  

But the inductor opposes the change in current by inducing an emf
  opposite to the direction of flow of current, isn't it?

Why?  If $I$ is increasing (decreasing), the voltage across the inductor $e$ is positive (negative) regardless of whether the instantaneous value of $I$ is negative or positive.
If you solve the differential equation for the current, you'll find that
$$I(t) = \frac{\mathcal{E}}{R}  + \left(I_0 - \frac{\mathcal{E}}{R}\right)e^{-tR/L}$$
where $I_0 = I(0)$ is the instantaneous value of the current when $t = 0$.
The voltage across the inductor is then
$$e = L\frac{dI}{dt} = \left(\mathcal{E} - I_0R\right)e^{-tR/L}$$
If you stare at that a bit, you'll see that voltage across the inductor is strictly positive for $I_0 < \frac{\mathcal{E}}{R}$ and strictly negative for $I_0 > \frac{\mathcal{E}}{R}$.
So, if the current is initially negative, the current will increase through zero to become positive (asymptotically approaching the steady state value $I_\infty = \frac{\mathcal{E}}{R}$) but the voltage across the inductor will always be positive.
In other words, $e$ will not change signs even though $I(t)$ does.
