For an 80% confidence interval, we would have a critical z value of

z_crit = norm.inv(0.9,0,1) = 1.281551566

The margin of error for the confidence interval will be z_crit x SEM

So 2.5 = 1.281551566 x sd/sqrt(n)

2.5 x sqrt(n) = 1.281551566 x sd

sqrt(n) = 1.281551566 x 5/2.5

n = 6.57

So, naively, we could say that the minimum required sample size is 7. In practice, we should take into account that we would be using the t-distribution, which would certainly increase the required sample size to 8-9, but since the degrees of freedom are dependent on n I'm not sure this is really what the example wants you to address (since it gets quite complex, mathematically).