Simplified form of the chemical equation.
Call BH the benzoic acid and B- the benzoate anion
BH = B- + H+
stoichiometry
Mo- x moles ---> xmol of B- and + xmol of H+
Where Mo is the initial concentration (before reaching the equilibrium) of BH and M0 - x is the concentration at equilibrium.
Ka = [x]^2 / [Mo - x]
[x] = [H+]
Unfortunately, It is not 100% clear if the 0.200M is the concentration at equilibrium or the initial concentration of BH.
I think it is the first. But else, you can do the numbers with the same procedure changing the denominator of Ka expression.
Then, ka = [H+]^2 / 0.200 = (3.55*10^-3)^2 / 0.200 = 0.0000630125 = 6.30*10^-5
Observe that if you suppose that the concentration of the solution at equillibrium is 0.2 - x instead 0.2, the result is very similar, because
ka = (3.55*10^-3)^2 / (0.2 - 0.00355) = 6.41 * 10^-5.
I still believe that the statement means that 0.200 is the concentration at equillibrium.
Then, my answer is 6.30*10 ^-5