Answer
$$
U=9.50 \mathrm{J} \text { }
$$
$$
d=0.396 \mathrm{m}
$$
Work Step by Step
The (final) elastic potential energy is
$$
U=\frac{1}{2} k x^{2}=\frac{1}{2}(431 \mathrm{N} / \mathrm{m})(0.210 \mathrm{m})^{2}=9.50 \mathrm{J} \text { . }
$$
eventually this must proceed from the source (gravitational) energy in the system $mgy$
(where we are measuring y from the lowest "elevation" reached by the block, so
$$
y=(d+x) \sin \left(30^{\circ}\right)
$$
Thus,
$$
\operatorname{mg}(d+x) \sin \left(30^{\circ}\right)=9.50 \mathrm{J} \quad \Rightarrow \quad d=0.396 \mathrm{m}
$$
