Answer: 62.5 N

Per Hooke's Law, in an ideal spring, the spring force F$_s$ is proportional to x where x is the displacement of the block from the equilibrium position

$F_s = -kx$.

$x = $ displacement of the block, i.,e, $x =0.75m - 0.5m = 0.25m$; It is given that the spring constant is 250Nm$^{-1}$

$\Rightarrow F_s = -k x = 250 Nm^{-1} \times 0.25\;m = 62.5 N$