The empirical rule says that about 68% of any normal distribution lies within one standard deviation of the mean. This leaves 32% of the distribution that lies outside this range, with about 16% to either side.
At the 16th percentile, there is a value of [tex]Y=y[/tex] such that
[tex]\mathbb P(Y<y)=\mathbb P\left(\dfrac{Y-470}\sigma<\dfrac{y-470}\sigma\right)=\mathbb P(Z<-1)\approx0.16[/tex]
where [tex]\sigma[/tex] is the standard deviation for the distribution, and [tex]Z[/tex] is the random variable corresponding to the standard normal distribution. This value of [tex]y[/tex] would correspond roughly to a z-score of [tex]Z=-1[/tex].
You're told that [tex]Y=340[/tex] lies at the 15th percentile, so that
[tex]\mathbb P(Y<340)=0.15[/tex]
Roughly, then, it'd be fair to say that [tex]y\approx340[/tex]. So you have
[tex]\dfrac{340-470}\sigma\approx-1\implies\sigma\approx130[/tex]
which falls between (A) and (B). To narrow down the choice, notice that [tex]y[/tex] would have be slightly larger than 340 in order to have [tex]\mathbb P(Y<y)=\mathbb P(Z<-1)\approx0.16[/tex]. This brings [tex]y[/tex] closer to the mean, and thus suggests the standard deviation for the distribution is actually smaller than our approximation.
This tells us that (A) is the answer.
