Assuming \(X\) is normally distributed, you have
\(\mathbb P(X>x')=\mathbb P\left(\dfrac{X-12}4>\dfrac{x'-12}4\right)=\mathbb P\left(Z>\dfrac{x'-12}4\right)=0.24\)
The right-tail probability of \(0.24\) corresponds to a \(z\) score of approximately \(0.7063\).
In terms of \(X\), this critical value is
\(0.7063=\dfrac{x'-12}4\implies x'\approx14.83\)
