Czarinah, this isn't meant to be a guessing game. You are missing information and conditions to obtain that answer; the most important piece being the size of the hole.

- Andrew

Salad, this sounds like a Fluids 1 problem, not sure I understand the issue with hole size? I doubt this question is intended to consider energy losses at the hole.

To simplify, I would convert potential energy at the hole, assuming the hole was recently put into the container and the water level has not yet dropped, and then convert to kinetic energy to solve for fluid velocity. This can be achieved using Bernouilli's equation, and is a classic question in basic fluid dynamic coursework.

@ChorasDen Yes, thank you. I did solved it using the Bernouillis equation..and came up with a velocity of 2 m/s. Is that reasonable?

Bernoulli euqation is the right way. Or you can use the simplified law of Toricelli, since the question only depends on the height of water surface to the hole.
Bernoulli -> Toricelli:
p1 + rho*g*h1 + 1/2*rho*v1^2 = p2 +rho*g*h2 + 1/2*rho*v2^2
p1 = p2 = patm
h1 = 0.2 m
h2 = 0 m

A1*v1 = A2*v2 -> A1 >> A2 -> v1 << v2 and v1 = 0 m/s

=>
rho*g*h1 = 1/2*rho*v2^2
-> sqrt(2*g*h1) = v2 (Toricelli)
with h1 = 0.2 m
-> v2 = sqrt(2*9.81*0.2) = 1.98 m/s