Answer
$U_{total} = \frac{1}{2}kx_1^2 + \frac{1}{2}(k+ka)(3x_1^2)$
Work Step by Step
We know that at $x_1$, the spring energy is equal to:
$=\frac{1}{2}kx_1^2 $
Since spring constants are additive when springs are in parallel, it follows:
$U_{total} = \frac{1}{2}kx_1^2 + \frac{1}{2}(k+ka)((2x_1)^2-x_1^2)$
$U_{total} = \frac{1}{2}kx_1^2 + \frac{1}{2}(k+ka)(3x_1^2)$
