According to the normal distribution table z=1.28 would put the score within about the top 10%, i.e., 90% of other scores would be lower. So we know that normalisation is z=(X-µ)/s where µ=mean (531) and s=standard deviation (74), so X=sz+µ=74*1.28+531=625.72. Round this up and we get 626 as the score at and above which would put the examinees in the top 10%.