Let [tex]Y[/tex] denote the random variable representing a given number in the total set of numbers. We're told that [tex]\dfrac{1360}{2000}=0.68[/tex] of the numbers fall within a given range, so we know
[tex]\mathbb P(40\le Y\le50)=0.68[/tex]
where [tex]Y[/tex] is normally distributed with mean 45 and an unknown variance [tex]x^2[/tex].
Let's make the transformation to a random variable with a standard normal distribution:
[tex]\mathbb P\left(\dfrac{40-45}x\le\dfrac{Y-45}x\le\dfrac{50-45}x\right)=\mathbb P\left(-\dfrac5x\le Z\le\dfrac5x\right)[/tex]
Since [tex]Z[/tex] is symmetric, we have
[tex]\mathbb P(-\dfrac5x\le Z\le\dfrac5x\right)=2\mathbb P\left(0\le Z\le\dfrac5x\right)=2\bigg(\mathbb P\left(Z\le\dfrac5x\right)-\mathbb P(Z\le0)\bigg)[/tex]
The mean of [tex]Z[/tex] is 0, and by symmetry we know that exactly half of the distribution falls to the left of [tex]Z=0[/tex], so [tex]\mathbb P(Z\le0)=0.5[/tex]. We're left with
[tex]0.68=2\mathbb P\left(Z\le\dfrac5x\right)-2\cdot0.5[/tex]
[tex]\implies\mathbb P\left(Z\le\dfrac5x\right)=0.84[/tex]
This probability corresponds to a value of [tex]Z\approx0.9945[/tex], which means
[tex]\dfrac5x\approx0.9945\implies x\approx5.0277[/tex]
