Suspended from a vertical spring, a block of mass 0.5 kg undergoes...

Angular frequency $\omega = \dfrac{2\pi}{T} = \dfrac{2 \times 3.14}{0.314} = 20 \text{ rad/s}$

Maximum acceleration of the block during SHM is a_\max = \omega^2 A = (20)^2 \times 0.1 = 40 \text{ m/s}^2

The maximum force exerted by the spring occurs at the lowest point of the oscillation where the spring is maximally stretched and the acceleration is directed upwards.

Using Newton's second law at the lowest point:

F_\max - mg = m a_\max

F_\max = m(g + a_\max)

Taking $g = 10 \text{ m/s}^2$:

F_\max = 0.5 \times (10 + 40) = 0.5 \times 50 = 25 \text{ N}