The maximum potential energy of a block executing simple harmonic motion is 25 J.  A is amplitude of oscillation. At A/2, the kinetic energy of the block is

The maximum potential in an S.H.M is given by 

$\frac{1}{2}{\mathrm{kA}}^2=25$

Potential energy of the simple harmonic oscillation at x is given by 

$\mathrm{U} = \frac{1}{2}{\mathrm{kx}}^2$

At x = A/2, U is

$\Rightarrow \frac{1}{2}\mathrm{K}{\left[\frac{\mathrm{A}}{2}\right]}^2=\frac{25}{4}$

We know total kinetic energy is given by T = K + U. 

Also, T is equal to the maximum potential energy. 

$\therefore \mathrm{K}+\mathrm{U}=25$

Therefore, at U = 25/4

$\Rightarrow \mathrm{K}=25-\frac{25}{4}=18.75\mathrm{J}$