In an exactly.200m solution of benzoic acid, a monoprotic acid, the [h+] = 3.55 × 10–3m. what is the ka for benzoic acid?

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

The answer is !