You're looking for [tex]k[/tex] such that
[tex]\mathbb P(X<k)=0.20[/tex]
Transforming the standard normal distribution, you have
[tex]\mathbb P(X<k)=\mathbb P\left(\dfrac{X-125}{12.51}<\dfrac{k-125}{12.51}\right)=\mathbb P(Z<k^*)[/tex]
where [tex]k^*[/tex] is the cutoff score in terms of this new distribution. The z-score corresponding to a probability of 0.20 is approximately [tex]z=k^*=-0.8416[/tex], which means the cutoff test score is
[tex]k^*=\dfrac{k-125}{12.51}\implies k=125+12.51k^*\approx114.47[/tex]
Rounding to the nearest whole point, a score of about 114 is the bare minimum requirement to not have to take the remedial course.
