Answer
The block rises to a point that is $~~0.39~m~~$ above the release point.
Work Step by Step
We can find the initial energy in the system:
$U_{e0} = \frac{1}{2}kd^2$
$U_{e0} = \frac{1}{2}(620~N/m)(0.25~m)^2$
$U_{e0} = 19.375~J$
Note that the block is not attached to the spring, so when the block reaches the highest point, there is no energy stored in the spring.
Therefore, at the highest point, $U_g$ is equal to the initial energy of $19.375~J$
We can use conservation of energy to find the maximum height $y$:
$U_g = 19.375~J$
$mgy = 19.375~J$
$y = \frac{19.375~J}{mg}$
$y = \frac{19.375~J}{50~N}$
$y = 0.39~m$
The block rises to a point that is $~~0.39~m~~$ above the release point.
