Voltage drops are greater than source voltage. How is it possible?
Because the voltage across the inductor is not DC, but a sinusoidal voltage, whose amplitude is 8V (or maybe 8V is its RMS value, it is not clear from the context).

That voltage is not in phase with the voltage across the resistor, so they don't add in the usual (DC) way.

You should use phasors (a vector-like representation of sinusoidal AC voltages) to perform the calculation exactly (using complex numbers, for example, or a graphical representation). 

In short, a sinusoidal quantity (voltage or current) is represented by a vector in the complex plane having a length which is equal to the amplitude of the quantity. The vector has its tail in the origin and it forms an angle with the positive real axis which is equal to the phase of the sine wave (with respect to some reference).

Usual Kirchhoff's laws holds also for phasors, but you need to use vector (or complex) addition instead of conventional real number addition.

In your circuit what you know is:


R and L are in series, so they share the same current
R and L are in series, so the vector sum of the voltages across them equals the overall voltage across the series (10V 50Hz).
the current is in phase with the voltage across R (that's a property of resistors)
the current has a phase lag of 90° with respect to the voltage across the inductor (that's a property of inductors)


Assuming the voltages shown in the schematic represent amplitudes and not RMS values (that's not very important since for a sine wave \$V_{amplitude} = \sqrt{2} \cdot V_{rms}\$), then you can draw a vector diagram where you have an horizontal vector (pointing right) of length 6V (the voltage across R) and a vertical vector (pointing downwards) of length 8 (the voltage across L).

Using the vector addition you find out that the voltage across the series has an amplitude which is the hypotenuse of a triangle having 6V and 8V as sides, hence:

$$
V_{across series} = \sqrt{6^2 + 8^2} = \sqrt{36 + 48} = \sqrt{100} = 10V
$$

Of course this is only the amplitude of the voltage. You can use trigonometry formulas to get the phase.

EDIT

This is what happens in the time domain:



the red curve is the voltage across the resistor (sine wave with 6V peak value), the blue one is the voltage across the inductor (sine wave with 8V peak value, delayed in time of a quarter of a period, which is what 90° phase lag means). If you add their value at any time, you get the green curve, which has a peak value of 10V and so it will never reach 14V.