Hey there Gnarls,
This question is phrased as if the diver hasn't fully decelerated by the time he reaches that depth, but we can only do this problem if we assume that the diver has slowed to a stop when they reach 1.85 m.
If we do, we can use this equation
vf2 = vi2 + 2aΔx
vf = 0m/s
vi = 17.64m/s
Δx = 1.85m
a = ?
Plug it all in we get
311.1696 m2/s2 = 0 + 2*a*(1.85 m) Remove 0
311.1696 m2/s2 =2*a*(1.85 m) Divide by 2
155.5848 m2/s2 =a*(1.85 m) Divide by 1.85 m
84.09989 m/s2 = a Round appropriately
So there you have it
The diver's acceleration was 84.10 m/s2