A 3 kg block is attached to a vertical spring. Initially, you exert a 50 N downwards force on the block, holding it in place, at rest. You let go. Find the instantaneous acceleration of the block immediately after you let go. What is the direction of the acceleration

As you were holding the block down and in place, the spring exerted an upward force that balanced the downward push by your hand and its own weight. So this restoring force has a magnitude of R such that

R - 50 N - (3 kg) g = 0   =>   R = 79.4 N

As soon as you remove your hand, the block has acceleration a such that, by Newton's second law,

R - (3 kg) g = (3 kg) a   =>   a = (79.4 N - (3 kg) g) / (3 kg) ≈ 16.7 m/s^2

pointing upward.

it may be wrong but i think its a