Let the speeds of the blocks A and B at the instant compression is $\frac{{\mathrm{x}}_0}{4}$ be v_A and v_B respectively as shown. (ℓ₀ is natural length)

Since, No external force acts on the system in Hz direction, therefore applying linear momentum conservation in Hz direction.
0 = m (–v_A) + 2m (v_B) or 
$\boxed{{\mathrm{v}}_{\mathrm{A}}=2{\mathrm{v}}_{\mathrm{B}}}$     ..............................(1)

Now, By conservation of energy,

$\boxed{\frac{1}{2}{\mathrm{kx}}_0^2=\frac{1}{2}\mathrm{k}{\left(\frac{{\mathrm{x}}_0}{4}\right)}^2+\frac{1}{2}\left(\mathrm{m}\right){\mathrm{v}}_{\mathrm{A}}^2+\frac{1}{2}\left(2\mathrm{m}\right){\mathrm{v}}_{\mathrm{B}}^2}$   ..................(2)

From (1) and (2) we get

$\frac{1}{2}{\mathrm{mv}}_{\mathrm{A}}^2=\frac{1}{2}{\mathrm{kx}}_0^2$

Using work energy theorem,
W.d. by spring on A = change in K.E. of A

$\text{ or }{\mathrm{W}}_{\mathrm{S}}=\frac{1}{2}{\mathrm{mv}}_{\mathrm{A}}^2=\frac{5}{16}{\mathrm{kx}}_0^2$