How do I calculate the standard deviation of a normal distribution given the mean and a quantile value of that distribution? How can I determine the standard deviation of a normal distribution with known mean and a known percentile value?  
The known percentile value would be on the correct side of the mean (eg, for percentile > 0.5, the value > mean), and the percentile would not be 0.5, as that would simply be equal to the mean, and therefore would tell one nothing about the variability.
This is an R function I use occasionally to do this; is there an analytic or arithmetic method/formula that could solve this? Would it be faster to compute?
#' Determine y in the equation `qnorm(mean=m,sd=y,p=p)==x`
get_sd_from_quantile_score <- function(m, p, x) {
  get_quantile_score <- function(y) { 
    qnorm(mean=m,sd=y,p=p)
  }
  f <- function(y) { (get_quantile_score(y) - x)^2 }
  opt <- optim(par = 1, fn = f, method = 'CG')
  return(opt$par)
}

 A: I believe simply (x-m)/qnorm(p) will do it, where m is the known mean and x is the known percentile p. qnorm(p) gives you how many standard deviations away from the mean the value is, and then you scale it by the given difference.
A: In general, you'll need to know two quantiles of a normal distribution
in order to determine $\mu$ and $\sigma.$ 
Here is an example based roughly
on the part of the Empirical Rule that says about 68% of the probability
under a normal curve lies between $\mu \pm \sigma.$ Thus quantiles .16 and .84
of $\mathsf{Norm}(\mu=100,\sigma=15)$ are roughly at 85 and 115. Computations in R:
qnorm(c(.16,.84), 100, 15)
[1]  85.08313 114.91687

pnorm(c(85, 115), 100, 15)
[1] 0.1586553 0.8413447

By formulating two probabilies with quantiles, standardizing, and using printed tables of the standard normal CDF, you can obtain and solve two equations in unknowns $\mu$ and $\sigma.$ I will leave details of this self-study problem to you. When you see how it goes, maybe you can evaluate your R functions.

Note: If the two quantiles are very close to each other, you may need the additional accuracy that R provides (above printed normal tables) to get accurate
values for $\mu$ and $\sigma.$
