The distribution being described here is normal, with a mean #mu = 85^o# and an unknown standard deviation #sigma#. Thus, we can describe this as #N(85^o,sigma^2)#.
To have 10% of days being at least (aka, greater than or equal to) #100^o#, we would need to find where on a standard normal distribution #N_S(0,1^2)# the z-score #z_{star}# would be such that the cumulative right-tail probability (the area under the standard normal curve to the right of #z_{star}#) is 0.10 (aka 10%).
One can find this using z-score tables or a z-score calculator. If you cannot find a z-score table that provides easy access for right-tail values, keep in mind that the standard normal distribution is symmetric about a z-score of 0, meaning you can find the equivalent cumulative left-tail z-score that provides an area of 0.10 (which will be a negative z-score value), and simply use the positive of that z-score.
In any case, the #z_star# we're looking for here is #z_star = 1.28#. This #z_star# value is on the standard normal distribution, and so to give a matching equivalent result for our original problem, we can use the z-score formula to work backwards from this z-score value to determine what the standard deviation would have to be for temperatures at least #100^o# to be in the 10% range:
#z = (x - mu)/sigma#
#1.28 = (100^o - 85^o)/sigma#
#1.28sigma = 15^o#
#sigma = 15^o/1.28 ~~ 11.72^o#
