How to take the reading of voltmeter if the needle falls between 1.6V and 1.7V?

So the resolution of the display is $pm 0.05$ it looks in the middle you you can state that $V=1.65pm 0.05 V$ So in general look at the gaps have half the length of the gap as your error and state where it is if it is closer the the bottom state in this example 1.6 or closer to the top 1.7 or in the middle 1.65 you are then covered by your error range.(manufacture documentation can help too )

That is a “mirror scale”: the silvery patch shows a reflection of the needle. To remove the parallax error, move your head so that the reflection appears directly under the needle. $newcommandX{text{X}}$

I was taught to mentally subdivide an analog scale into ten divisions. It’s pretty easy to decide whether a needle is “on the line” (in which case, $X.0$, where $X$ is the value of the mark) or “in the middle” (in which case $X.5$). If the needle is in the lower half of the gap but not on $X.0$ or $X.5$, I call it either “near the quarter,” in which case it’s $X.2$ or $X.3$, of “far from the quarter,” $X.1$ or $X.4$. If I can’t decide on a least significant digit, then I just pick one: that’s the whole point of having an error bar. If I am making many measurements and frequently can’t decide, then I quantify this by making all of my error bars bigger.

Using this technique with your photo,

reading where the needle interacts with the marks, I prefer 1.57. (Both 1.55 and 1.59 are clearly wrong. The needle is beyond the center, but could get closer to the next line without meeting it.) But 1.57V is not the value on the meter, because of the parallax error: the reflection is to the right of the needle. Look again, from the right angle; we can’t read your meter from this photograph.

If you cannot re-measure without your parallax error, and have only this photo: you know that the projection of the needle onto the plane of the paper where the marks are is going to lie between the apparent position of the need and the apparent position of its reflection. The needle is definitely above 1.55V, and the reflection is probably below 1.70V, so a dumb average is 1.62V. Call it $(1.62pm 0.03)text V$ to account for the extra uncertainty in undoing the parallax.

The proper thing is just to incorporate the parallax error into the measurement error. You know that “doing it right,” you would shift your head so that the two overlap, and you could plausibly make out uncertainties to maybe a fifth of a tick, so you'd maybe report some measurement $pm 0.01text{ V}$ or so, unless the needle had a jitter to it or some other thing that forced you to not be able to quite get that precision.

But, this measurement was not done perfectly and now has an additional source of error. Nevertheless we can kind of reason that probably the needle is a lower bound and the reflection is an upper bound for the measurement, because parallax would tend to shift the needle right relative to the scale ticks, and the reflection left, until they met in the middle. So the measurement is now telling you it is between 1.56 or so, as an estimate of how low the needle could be, and 1.64, as an estimate of how high the reflection could be. As a result I would probably say something like $1.60pm0.03text{V},$ the uncertainty created by the parallax error is maybe three times the size of the uncertainty intrinsic to the device.

Uncertainty is not something that is magical or objective. It's like in this word certain, we are intrinsically talking about a subjective property that has to do with you and how you feel about things. Of course we need you to be honest and realistic in your feelings, otherwise people will not trust your estimates of your uncertainty for estimates of their own uncertainty and they will have to do their own estimation... But yeah, that's what uncertainty is about. It's not about an exact number that a machine will spit out at you, it's about affirming all such machines (and ourselves and our process) as imperfect. For this number you could want to quote $1.60$ or maybe I might skew slightly right to $1.61$ for the measurement, but with much larger error bars because of this parallax error.