Answer
The high-tension string has about $7\%$ more tension.
Work Step by Step
Relate the frequency f, length L, tension $F_T$, and mass per unit length, $\mu$.
$$f=\frac{v}{2L}=\frac{1}{2L}\sqrt{\frac{F_T}{\mu}}$$
The mass per unit length is the density multiplied by the cross-sectional area.
$$f=\frac{1}{2L}\sqrt{\frac{F_T}{\rho\pi r^2}}$$
Solve for the tension.
$$F_T=4L^2\rho f^2\pi r^2$$
Take the ratio of high (H) to low (L) tensions.
$$\frac{F_{T,high}}{F_{T,low}}=\frac{4L^2\rho f^2\pi r_H^2}{4L^2\rho f^2\pi r_L^2}$$
The 2 strings are tuned to the same frequency, they have the same length, and the nylon density is the same.
$$\frac{F_{T,high}}{F_{T,low}}=\frac{ r_H^2}{ r_L^2}=\frac{ d_H^2}{ d_L^2}$$
$$\frac{F_{T,high}}{F_{T,low}}=\frac{ (0.724mm)^2}{ (0.699mm)^2}=1.07$$
The high-tension string has about $7\%$ more tension.
