Answer
$\approx 1.4\;\rm cm$
Work Step by Step
Assuming that the dimensions of the loop are much smaller than 50 cm, so the on-axis magnetic field strength at 50 cm is given by
$$ B =\dfrac{2\mu_0 \mu }{4\pi z^3}$$
where $\mu=AI$ and $A$ is the cross-sectional area of the loop which is a square, so $A=L^2$.
$$ B =\dfrac{2\mu_0IA }{4\pi z^3} =\dfrac{2\mu_0IL^2 }{4\pi z^3}$$
Solving for $L$;
$$L=\sqrt{\dfrac{4\pi z^3B}{2\mu_0I}}$$
Plug the known;
$$L=\sqrt{\dfrac{4\pi (0.5)^3(7.5\times 10^{-9})}{2(4\pi \times 10^{-7})(25)}}$$
$$L=\color{red}{\bf 1.37}\;\rm cm$$
