Quantization error is the difference between the actual input and the level the converter reports. If each code represents the centre of its interval, the worst error occurs at either edge of the interval and equals half its width:

\(e_{max}=\dfrac{\Delta}{2}\)

First check that the intervals add up.

\(2\times\dfrac{V}{14}+6\times\dfrac{V}{7}=\dfrac{V}{7}+\dfrac{6V}{7}=V\ \checkmark\)

so the eight intervals do span the full range.

Now take the largest interval, which is one of the six middle ones of width V/7:

\(e_{max}=\dfrac{1}{2}\times\dfrac{V}{7}=\dfrac{V}{14}\)

which is option 3.

Why the end intervals were made half-width — this is the whole design idea. The two extreme intervals have their code at the edge of the range (0 and V) rather than at a centre, so an error of half a full interval is impossible there; making them V/14 wide gives them a maximum error of exactly V/14 as well. Every interval in the converter therefore contributes the same worst-case error:

This is precisely the mid-tread arrangement used in real converters, and it is why a datasheet quotes \(\pm\tfrac{1}{2}\) LSB rather than \(\pm1\) LSB.

Compare with the naive uniform design. Eight equal intervals of V/8 with codes at the interval boundaries would give a maximum error of V/8 at some points — worse than V/14 — and the error would not be symmetric about zero. Splitting the end steps buys a uniform, symmetric error at no extra hardware cost.

The general result for an n-bit converter of full scale V is

\(e_{max}=\dfrac{V}{2\left(2^{n}-1\right)}\)

and with n = 3 this gives \(\dfrac{V}{2\times7}=\dfrac{V}{14}\), confirming the answer. Quantization noise power is then \(\Delta^{2}/12\), giving the familiar 6 dB of signal-to-noise ratio per bit.

Hence, the maximum quantization error is V/14.